Cambridge International
AS & A Level Mathematics
Probability & Statistics 1
STUDENT’S BOOK: Worked solutions
Louise Ackroyd, Jonny Griffiths, Yimeng Gu
Series Editor: Dr Adam Boddison
Pure Mathematics 1 International Students Book Title page.indd 1
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1
WORKED SOLUTIONS
Worked Solutions
1 Representation of data
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication
of the answer.
Prerequisite knowledge
Exercise 1.1A
1
1
a The data is quantitative, as it is numerical, and
discrete, as there are distinct individual values.
(The original data could have been continuous,
and then rounded to the nearest mark.)
b Monthly texts you send on your mobile is
discrete, as you can count them.
b Median = 35, mode = 41 and range = 35
c
Grouped scores
c The number of burgers sold in a fast food
restaurant is discrete, as you can count
them.
Frequency
10–19
6
20–29
13
30–39
13
40–49
18
d The duration of a marathon is continuous,
as you measure it.
A histogram with equal class widths is a
sensible way to display the data.
Frequency
2
20
c An IGCSE grade in maths as a letter is
qualitative.
10
5
19.5
29.5
39.5
d Examination scores in maths are quantitative
and discrete.
49.5
Grouped scores
e Waist size is quantitative and continuous,
unless it is given to the nearest cm/inch, in
which case it is discrete.
The results show a modal group of 40–49.
2
3
e The age of a teacher in your school is
continuous, as time is measured. But if
someone is asked their age, they reply with
a whole number, and that version of ‘age’ is
discrete.
a Gender is qualitative.
b Height is quantitative, as it has a numerical
meaning and is continuous.
15
0
9.5
a Daily rainfall in Penang is continuous, as you
measure it.
Edam: 90 × 120 = 30 , Stilton: 102 × 120 = 34,
360
360
168 × 120 = 56
Cheddar: 360
f ‘Car owner or not’ is qualitative.
Score
1
2
3
4
5
6
Frequency
3
4
2
3
3
5
Mode = most common value = 6
Mean = sum of all the values divided by 20 = 74 = 3.7
20
Median = middle value when results are ordered = 4
Range = top result − bottom result = 6 − 1 = 5
3
g Weekly self-study time is quantitative and
continuous unless measured to the nearest
hour, in which case it is discrete.
a 9
b 9
c 13.05
d The mean takes into account all the visits
made, so it is the best measure to use.
1
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1 Representation of data
4
5
a Farah’s first week sums to 22.4 mm.
22.4 + 84.6 = 107, and 107 = 3.82 mm
28
b No effect, since 9.4 + 0.5 = 4.9 + 5.0.
Number of
times stopped
Number of
journeys
10 a Median = 39th value = 3 years
Mode = 2 years
Mean = 276 ÷ 77 = 3.58 years
b The quartiles LQ, median and UQ divide the
ordered data into four equal sets.
fx
1
3
3
2
5
10
3
11
33
4
21
84
5
22
110
6
17
102
7
14
98
8
7
56
Total
100
496
LQ = 19.5th value = 2, UQ = 58.5th value = 5,
IQR = 3
c
1
2
3
4
5
6
7
8
9
10
Exercise 1.2A
1
Range = 41, Q1 = 20, Q3 = 44.5, so IQR = 24.5
2
∑ x = 1705,∑ x 2 = 537 615 so
( )
537 615 1705 2
var(
x
)
=
−
= 22 920 (4 s.f.), s.d.(x) = 151 (3 s.f.)
+
100
1
= 50.5th value, so 5
Median =
15
15
2
537 615 1705 2
var(x) =
−
= 22 920 (4 s.f.), s.d.(x) = 151 (3 s.f.)
15
15
Mode = 5
( )
3
Mean = 496 = 4.96
100
6
he best average to represent the data is the mean,
T
as it uses all the data, which contains no outliers.
aMean = ∑ f × mid-point and so
∑f
IQR = 2, s.d. = 4
4
∑ x = 191,∑ x 2 = 2275
b Since all the mid-points increase by a factor of
1.1, the mean increases by a factor of 1.1.
2
There is an argument for the IQR, since this is
unaffected by the high value of 20. On the other
hand, the standard deviation does utilise all the data.
d a = 0.1 × 19.3 = 1.93
Height, h (cm)
Frequency
fx
100 < h  120
5
550
120 < h  140
4
520
Mean = ∑ xi = 3985.1 = 332.091... = 332.1 cm 3
12
12
2
2
x
2
Variance = ∑ i − x = 1 323 540.07 − 3985.1
12
12
n
140 < h  160
12
1800
= 10.1307… and so standard deviation = 3.18 cm3
160 < h  180
13
2210
180 < h  200
8
1520
Total
42
6600
Estimate of the mean is correct: 6600 = 157.14
42
9
( ) = 3.71
2275 − 191
18
18
σ=
s.d.(x)
c Since all the mid-points increase by a, the mean
increases by a.
8
Range = 20 – 5 = 15
IQR = 14 – 8 = 6
6187.5 ÷ 320 = 19.335… = 19.3
7
For example, −a, −a, −1, −1, −1, 0, 1, 1, 1, a, a
where a = 42.5 = 6.52 (3 s.f.)
24 × 84.1 + 4 × 240.3 + 51 × 99.2 = 101.8 kg f × mid-point
79
iAnswers to a, c and d are qualitative; answers to
b, e and f are quantitative.
ii
Answer to b is discrete; answers to e and f are
continuous.
iii Check students' answers
5
(
6
)
Method 1
2
2
2
 1 − 31  +  4 − 31  +  5 − 31  +  6 − 31 

5  
5  
5 
5  
2
2
+  15 − 31  = 110.8

5
s.d.(x)
σ = 110.8 = 4.71 (3 s.f.)
5
Method 2
( ) = 4.71 (3 s.f.)
− 31
∑ x = 31, ∑ x 2 = 303, s.d.(x) = 303
5
5
2
2
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Variance of combined
samples
WORKED
SOLUTIONS
(2)
= n(b + a ) + n(a + b ) − ( a + b )
2n
2
=
7
∑b = 64.4, ∑ g = 68.6 ⇒ ∑b = 1030.4, ∑ g = 891.8
16
∑ b 2 − 64.42 ⇒ ∑ b 2 = 66 520.96,
16
∑ g 2 − 68.62 ⇒ ∑ g 2 = 61 295.78
9.1 =
13
⇒ ∑ b 2, g 2 = 127 816.74
⇒ var(b, g ) =
8
(
)
127 816.74 1922.2 2
−
= 14.1 (3 s.f.)
29
29
aThis is FALSE. The variance could be 0.64, in
which case the standard deviation would be 0.8.
b This is TRUE. The range tells you the span of
all the data, while the IQR tells you where the
central 50% of the data lies.
c This is FALSE. Given a set of data, we can
subtract the same number from every member
of the set to create a new set with a smaller
mean but with an unchanged variance.
d This is FALSE. The data set 1, 2, 3, 4, 5 has
IQR = 3 and variance = 2.
9 L
et x be the number of sweets in a bag sampled on
the first day, and y the number of sweets in a bag
sampled on the second day.
a ∑ x = 300,∑ y = 360,∑ x 2 = 3360,∑ y 2 = 4620
So mean(x, y) = 11, var(x, y) = 12
b In general ∑ x = na,∑ y = nb,∑ x 2 = n(b + a 2),
∑ y 2 = n(a + b 2)
So mean (x, y) = a + b , var(x, y) = a + b + (a − b)
2
4
2
2
(
a
−
b
)
so var(x, y) – mean(x, y) =
0
4
2
For the combined samples,
variance – mean =
10 aThe range has the advantage that it is easy to
calculate, but the disadvantage that it is affected
significantly by extreme values. The company
may have a few very highly paid people that
could distort the picture.
bThe interquartile range is more sensible than
the range as a measure of spread since it gives
you the range for the central 50% of the data,
so is unaffected by extreme values. However,
in concentrating on the central 50%, it has the
disadvantage of ignoring 50% of the data.
cThe variance includes all the data in its
calculation, which is an advantage, but extreme
values can still have a distorting impact.
dYou can say the same for the standard deviation
as the variance, but one advantage of the s.d.
over the variance is that its units are the same as
for the data.
11 aShe needs to score 74.
b
24, 34, 37, 39, 42, 54, a
20
She needs to score anything more than 54.
2
9302 + a 2 −  230 + a  = 112 ⇒ a = 57
c
 7 
7
She needs to score 57.
∑ x 2 − a 2 = b ⇒ ∑ x 2 = n(b + a 2)
n
∑ y 2 − b 2 = a ⇒ ∑ x 2 = n(a + b 2)
2
x + ∑ y na + nb a + b
=
=
2n
2n
2
Variance of combined samples
(2)
= n(b + a ) + n(a + b ) − ( a + b )
2n
2
∑x2 + ∑ y2 − a + b 2
2n
2
2
2
(
)
9302 + a 2 −  230 + a  = 10 2 2 ⇒ a = 71
d
 7 
7
She needs to score 71.
n
=
2
a + b + (a − b)2 − a + b = (a − b)2 0
2
4
2
4
so
variance
mean.
Then ∑ x = na, ∑ y = nb
Combined mean = ∑
2
Let x represent data values from the first
sample, and y data values from the second
sample.
2n
2
2
2
2
2
= b + a + a + b − a + ab + b
2
4
2
− 2ab − b 2
2
2 2 2
2 2
= b+ a + a+ b −a
4
2
b
+
a
a
−
ab
+ b2
2
2
2
=
+
4
4
2
= a + b + (a − b)
2
4
⇒ ∑ b, g = 1922.2 ⇒ b, g = 66.3 (3sf
. .)
10.2 =
2
2
13
1
∑x + ∑ y − a + b 2
2
Exercise 1.3A
1
Various ideas such as height, weight, number of
texts sent, number of music tracks owned, etc.
Discrete variables are counted, while
continuous data is measured.
2
2
2
2
= b + a + a + b − a + 2ab + b
2
4
3
2
2
2
2
2b −Cambridge
a − 2ab −International
b
©HarperCollinsPublishers
AS & A Level Mathematics: Probability & Statistics 1 9780008257767
= 2b + 2a + 2a + 2018
4
2
2
= 2b + 2a + a − 2ab + b
4
4
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a + b (a − b)2
1 REPRESENTATION OF DATA
2
b Median is similar but Molly’s is slightly higher,
suggesting that she scores more. Larger range for
Molly, suggesting that Jenson is more consistent
overall. On the other hand, Jenson has a slightly
larger IQR, suggesting that Molly’s middle 50% of
scores are more consistent.
For a histogram, the frequency a bar represents is
proportional to its area.
Since 47 goes with an area of 8 cm2, one bean is
represented by 8 cm2.
47
Since data is measured to the nearest cm, the
true class intervals are:
5
2.5–8.5, 8.5–13.5, 13.5–25.5
b Q0: 36, Q1: 46, Q2: 56, Q3: 64, Q4: 73
These have widths of 6, 5, and 12 respectively.
(IQR = 18)
The width of 6 goes with the given width of
2 cm, so the other classes have widths 5 cm
3
and 4 cm. The 22 beans in the second class are
represented by an area of 8 × 22 cm2 = 3.75,
47
and dividing by 5 gives a height of 2.25 cm.
3
5
a Width = cm
3
b Height = 2.25 cm
c
30
6
a We can treat the data as being continuous
here. Grouping the data means a histogram
is appropriate.
b Because it includes both start and end values.
c
True distance
(km)
Number of
people
40
50
60
70
80
Test mark
Price of
printer ($)
Frequency
Class
width
Frequency
density
50 < x  70
24
20
1.20
70 < x  100
31
30
1.03
100 < x  130
42
30
1.40
Frequency
density
130 < x  160
16
30
0.53
160 < x  200
18
40
0.45
200 < x  300
4
100
0.04
39.5–49.5
67
67 = 6.70
10
49.5–59.5
124
124 = 12.40
10
59.5–64.5
4023
4023 = 804.6
5
64.5–69.5
2981
2981 = 596.2
5
69.5–84.5
89
89 = 5.93
15
84.5–149.5
75
75 = 1.15
65
1.4
Frequency density
3
a Mode = 60
1.2
1.0
0.8
0.6
0.4
0.2
50
100
150
200
250
300
Printer cost ($)
4
a Jenson Q0: 3, Q1: 12, Q2: 31, Q3: 55, Q4: 66
Molly
The histogram has a hump to the left and a tail to
the right. Most of the printers cost less than $150,
but the small number of more expensive printers
means the mean price will be above the median
price.
Q0: 12, Q1: 17, Q2: 34, Q3: 57, Q4: 98
Jenson
Molly
7
a i True
ii True
iii False. The IQR for B is smaller than that for A.
0
10 20 30 40 50 60 70 80 90 100
iv False. Q3 − Q2 is larger for B.
Number of runs
4
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1
WORKED SOLUTIONS
8
b There is a larger range for Class B than for
Class A. The IQR for B is smaller than that for
A, however, so the middle 50% in class B is
more tightly grouped. The medians for the
two student groups are close, suggesting that
attainment is roughly equal for the two groups.
a
Exercise 1.4A
1
Number in
family
Number of
families
Cumulative
frequency
1
15
15
2
20
35
3
22
57
4
23
80
Class Frequency
width
density
Mass of
raisin (g)
Frequency
0 < x  0.5
4
0.5
8
5
11
91
0.5 < x  1.0
8
0.5
16
6
4
95
1.0 < x  2.0
14
1
14
2.0 < x  3.0
6
1
6
3.0 < x  3.5
4
0.5
8
Median = 48th value, and is approximately 2.6
from the cumulative frequency diagram.
Cumulative frequency
Frequency density
15
10
5
0.5
1
1.5
2
2.5
3
65
88776420
955333310
55420
630
70
60
×
50
40
×
30
20
0
×
Median = 2.6
1
2
3
After
0
1
2
3
4
5
5589
123566899
0004467789
089
Key: 4 2 3 = 24 marks before,
0
23 marks after
The median mark for the group has risen from
23 to 30; the lesson seems to have had an effect
(although there could be other reasons for the
increase).
4
5
6
Number in family
b The 0.5–1.0 class
Before
×
×
80
10
3.5
Mass of raisin (g)
9
×
90
20
2
Kick distance, l (m)
Upper
bound
Cumulative
frequency
5  l < 10
10
5
10  l < 20
20
58
20  l < 30
30
87
30  l < 50
50
102
50  l < 70
70
113
70  l < 100
100
120
10 You have Q1 − 1.5 × (Q3 − Q1) = 3,
Q3 + 1.5 × (Q3−Q1) = 83
Rearranging gives 5Q1 − 3Q3 = 6, 5Q3 − 3Q1 = 166
Solving these simultaneously gives Q1 = 33, Q3 = 53
5
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1 REPRESENTATION OF DATA
a
b a = 15, b = 85, central 70% of results lie in
(31,72).
c Mean = ∑
120
Cumulative frequency
90
80
4
70
b Q1: 46, Q2: 56, Q3: 64
60
(IQR = 18)
50
c
40
50
30
45
Cumulative frequency
P40 = 18
10
0
10 20 30 40 50 60 70 80 90 100 110 120
Kick distance, l (m)
b 40 × 120 = 48th value ≈ 18
100
Kick
distance,
l (m)
Midpoint
Frequency
5  l < 10
7.5
5
37.5
10  l < 20
15
53
795
20  l < 30
25
29
725
30  l < 50
40
15
600
50  l < 70
60
11
660
70  l < 100
85
7
595
Total
120
3412.5
fx
40
35
30
25
20
15
10
5
0
Mean ≈ 3412.5 ≈ 28.4
120
30 35 40 45 50 55 60 65 70 75 80
Test mark
d P90 − P10 = 69 – 41 = 28 marks. This measure
is useful because it omits extreme values.
5
Number of
goals
0
1
3
4
5
6
7
8
Frequency
7
9 15 9
6
7
4
4
2
Cumulative
frequency
7 16 31 40 46 53 57 61 63
a Median ≈ 48
Cumulative frequency
850
500
50
40
30
20
10
1
150
0
2
60
1000
Cumulative frequency
50 855
= 50.9
1000
a Mode = 60
20
3
=
The mean and median are similar, so the
distribution is fairly symmetrical.
100
c
fx
∑f
110
50
100
2
3
4
5
Number of goals
6
7
8
The median value is slightly more than 2 goals
per game.
Number of aphids
6
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1
WORKED SOLUTIONS
a
Upper class values
Cumulative
frequency
18.0
15
18.5
42
19.0
60
20.0
72
25.0
87
30.0
91
90
Cumulative frequency
6
80
70
60
50
40
30
20
10
b
25
Cumulative frequency
100
90
80
You need P15 and P85, and
70
50
40
30
15 × 82 = 12.3,
100
P85 – P15 is approximately 29.7 – 27.6 =
2.1 seconds.
Median = 18.6cm
20
18
9
20
22
24
26
28
30
Length (cm)
c Median ≈ 18.6 cm
d Mean =
∑ f × mid-point = 1773.45 = 19.5 (3 s.f.)
91
∑f
Thus the mean is greater than the median,
since the bulk of the results are to the left of the
distribution, with a long tail to the right.
The median is the value of x at which y = 320:
300 − 80 = 5.5
40
25 × 600 = 150, 75 × 600 = 450
, so
100
100
450 = 22.25.
150 − 80 =
1.75, Q3 =
Q1 =
20
40
Thus the interquartile range is 20.5.
a 63 minutes
c Q3 − Q1 = 80 − 46 = 34 minutes
d Approximately 42 minutes
e Cannot tell using just the box plot.
1.6 hours is 1 h 36 minutes. Using the
cumulative frequency curve, we can draw up
from 96 minutes to the curve, and then read
off approximately 175 on the vertical axis.
Exercise 1.5A
1
b
f Box plot is more compact, but only shows
five values (the quartiles), so the cumulative
frequency curve shows more information.
Time (seconds) 26 27 28 29 30 31 32 33+
Cumulative
frequency
1
1
3 15 35 23 4
1
a
∑ x = 56, ∑ x 2 = 560 , mean = 8,
standard deviation = 4
This suggests that about 25 sessions out of
the 200 last longer than 96 minutes.
Frequency
Only a and f could be parts of a genuine
cumulative frequency diagram.
A cumulative frequency diagram can never be
decreasing, which rules out b, c and d.
A frequency can never be negative, which rules
out e.
10 The maximum value for y is 600, and the minimum
value is zero. The curve is never decreasing and
starts with a y-value of zero, so it can be a complete
cumulative frequency curve.
b 140 minutes
8
35
85 × 82 = 69.7
100
60
10
7
30
Time (seconds)
0
4 19 54 77 81 82 82
∑ x = 350,∑ x 2 = 25 500, mean = 70,
standard deviation = 14.1 (3 s.f.)
c
2
∑ x = 91, ∑ x 2 = 1275 , mean = 13,
standard deviation = 3.63
a
∑ x = 50, ∑ x 2 = 298 , mean = 5,
standard deviation = 2.19
b ∑ x = 115.8, ∑ x 2 = 1981.58 , mean = 16.5,
standard deviation = 3.07
7
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1 Representation of data
3
a
∑ fx = 194,∑ fx 2 = 1216 ,
8
mean = ∑
fx
= 6.06
∑f
x = 2641 = 66.025 = 66.0 (3 s.f.),var(x) =
40
2
s.d.(x) =
∑ fx 2 −  ∑ fx  = 1.12
∑ f  ∑ f 
bThe smallest possible variance is 0, which
would mean
∑ x 2 − 66.0252 = 0 ⇒ ∑ x 2 = 174 372.025
All three averages are similar, which is due to
the symmetrical nature of the distribution.
However, the median or mode is a more
useful value because it is an actual shoe size.
∑ x = 27, ∑ x = 245, n = 13
2
5
13
 13 
2
2
Standard deviation = 1181 −  102  = 5.70 (3 s.f.)
15  15 
− = x − 10, s.d.(w) = s.d.(x), so
aIf w = x – 10, then w
b
200
∑ fm 2 − 73.6752 = 1110 737.5 − 73.6752 = 125.682...
200
200
Since there is now less variation in the masses of
the apples, the standard deviation of the masses
of the new batch will be slightly reduced.
10 ∑(v − 127.5) ≈ 19 × 0 + 24 × 10 + 35 × 20 + 41 × 30
+ 27 × 40 + 9 × 50 = 3700
∑(v − 127.5)2 ≈ 19 × 02 + 24 × 102 + 35 × 202 + 41 × 302
∑(x − 8) = 10.6 − 8 ⇒ ∑(x − 8) = 52,
+ 27 × 40 2 + 9 × 50 2 = 119 000
20
∑(x − 8)2 − 2.62 ⇒ ∑(x − 8)2 = 272 (3 s.f.)
20
y = 48 + a = 10.6 ⇒ a = 9
30
( )
2
var(y) = 314 − 48 = 7.91 (3 s.f.) (So in fact we
30
30
do not need to know the value of a here).
7
∑ fm = 770 + 1750 + 2092.5 + 3335 + 3487.5 + 2040 + 1260
x = 10 + 12 = 10.6, s.d.(x) = 144 − 0.6 2 = 2.62
20
20
6.84 =
c
a Mean
⇒ standard deviation
⇒ σ = 11.2 (3 s.f.)
bSince the apple has the mean mass, the mean of
the new batch of apples will be unchanged.
x
Mean = ∑ = 102 = 6.8
n
15
1181  102 
Variance = 15 −  15  = 32.5 (3 s.f.)
6
x
∑ 2 must be a whole number, so the smallest it
can be is 174 373.
200
= 73.675 = 73.7 (3 s.f.)
∑ x 2 −  ∑ x  = 245 −  27  2 = 3.81
 n 


40
9
x
Mean = ∑ = 27 = 2.08
n
13
Standard deviation =
n
175 042
− 66.0252 = 16.749... ⇒ s.d.(x) = 4.09 (3 s.f.)
40
− 66.0252 = 16.749... ⇒ s.d.(x) = 4.09 (3 s.f.)
b Both median and mode = 6.
4
175 042
2641
a Mean x = 40 = 66.025 = 66.0 (3 s.f.),var(x) = 40
a t = 20, s.d.(t ) = 4.60, c = 20, s.d.(c) = 1.41
b The country route, as there is a smaller
deviation so it is easier to predict how long it
will take, and is no slower on this evidence.
Mean ≈
3700 + 127.5 = 151 (3 s.f.)
155
Variance ≈
2
119 000  3700 
−
= 198 (3 s.f.)
 155 
155
11 If y = x – 25, then y = x − 25, s.d.(y) = s.d.(x).
y = 124 = 6.2, s.d.(y) = 11.8. This implies
20
x = 31.2, s.d.(x) = 11.8.
12 If y = x – 70, then y = x − 70, s.d.(y) = s.d.(x).
y = −315 = − 6.3, s.d.(y) = 3.41. This implies
50
x = 63.7, s.d.(x) = 3.41 (3 s. f.)
2
c
2106 + p 2  100 + p 
−
= 20, which becomes
6
 6 
2
5p – 200p + 1916 = 0
This solves to give p = 16 or 24 (to the nearest
integer).
8
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Variance =
∑ fx 2 −  ∑ fx 
∑ f  ∑ f 
2
(
)
WORKED
6 094
779 705 SOLUTIONS
1 263 541
=
−
1069
1069
= 4 304 297.79
Exam-style questions
a
≈ $2075 (to the nearest dollar)
6 a The mid-points are 3 and 25.5 respectively.
b 7 students are represented by an area of 10 cm2, so
1 student is represented by an area of 10 cm2. The
7
true class width of the 6–10 class is 5 hours, and
the true class width of the 31–49 class is 19, so the
width of the 31–49 class on the histogram will be
19
= 7.6 cm. The height of the 31–49 class will be
2.5
14 5 5 6 6 8 8
15 0 1 1 2 3 5 6
16 0 1 1 2 4
17 0 0
Key: 14 5 = 1.45 m
b
1.4
1.5
1.6
10
7
19 = 0.188 cm.
2.5
1.7
Height (m)
2
c Estimate of the mean = ∑
a Median = Q2 = 187, Q1 = 121, Q3 = 260
7
b IQR = 139, Q1 − 1.5 × IQR = −87.5,
Q3 + 1.5 × IQR = 468.5
All the 11 values are inside the range (−87.5,
468.5), so there are no outliers in the data.
3
a Boys
b 13
14
15
16
17
18
9
322
6322
952
63
44
4
2 3 7 8 9
2 5 6
6 6 7
2 3 9
13
14
15
16
17
18
4
Girls
23789
256
667
239
Key: 3 14 2 = boys 143,
girls 142
Girls
5
t
5–10 10–14 14–18 18–25 25–40
Frequency
10
15
22
21
18
Frequency
density
2
3.75
5.5
3
1.2
100
90
80
70
60
50
40
30
10
∑ x = 2392 for current boys. For boys with
new boy added ∑ x = 160 × 60 = 2560.
0
5
10 15 20 25 30 35 40 45
t (minutes)
b Approximately 86 − 53 = 33 people
Thus new boy’s score is 168.
4
a
fx 602
=
= 15.05 hours
∑ f 40
20
IQR for the girls = 167 − 147 = 20
c
1
s.d. = 4304297.79 = 2074.68
Cumulative frequency
1
2
100 + 2(7) = 114 cm, and 100 − 2(7) = 86 cm.
c Mean = ∑
So if the rainfall in the city is outside the range
(86, 114) the year can be taken as exceptional.
d Standard deviation ≈
aMean profit = ∑
fx 1 263 541
= $1182
=
1069
∑f
(to the nearest dollar)
b Variance = ∑
fx 2  ∑ fx 
−
∑ f  ∑ f 
=
2
(
6 094 779 705 1 263 541
−
1069
1069
= 4 304 297.79
fx 1643.5
= 19.1 (3 s.f.)
=
86
∑f
∑ fx 2 − x 2 = 37 074.25 −  1643.5  2
 86 
86
∑f
= 8.12 minutes (3 s.f.)
e Q1 ≈ 13
)
Median = Q2 ≈ 18
2
Q3 ≈ 23, P60 = 18 minutes
s.d. = 4304297.79 = 2074.68
≈ $2075 (to the nearest dollar)
9
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1 REPRESENTATION OF DATA
8
a The median is the 123rd result, which lies in the
170–180 class.
10 a
Cumulative frequency
90
b
Frequency density
8
7
6
5
4
70
60
40
30
20
10
2
0
Median ≈ 20.8
×
×
×
12 14 16 18 20 22 24 26
Foot length, l (cm)
b ∑ fx = 2077,∑ fx 2 = 43524.5
Length of song (sec)
Mean ≈ 20.8 cm
fx
c Mean ≈ ∑ = 42445 = 173 sec (3 s.f.)
246
∑f
Standard deviation ≈ 1.96 cm (3 s.f.)
a The class widths are 200, 40, 40, 70, 150, 300, so
the frequency densities are 0.03, 0.5, 0.75, 0.343,
0.08, 0.0267.
The area of the 200–240 class is 14, so the
area per household is 0.7, so the area for
the 280–350 class is 16.8 cm2.
The width of the 280–350 class is 2 × 70 =
40
3.5 cm, so its height must be 4.8 cm.
100
×
90
c Mean is best as no obvious outliers, and it
uses all available data.
11 a Mean = 146 +160 = $164.87
30
2
Variance = 1024 − 146 ⇒ standard deviation = $3.23
30
30
b
∑ x 2 − 320∑ x + 30 × 1602 ⇒ ∑ x 2 = 815 744
60
×
50
c Mean = 4946 = 164.9 (4 s.f.)
30
Variance = 815 744 − 164.8666 2
30
⇒ standard deviation = $3.23 (3 s.f.)
40
30
×
20
×
0
∑ x = 146 + 160 × 30 = 4946,∑ x 2 = 30
Alternative method: multiplying out, 1024 =
70
10
2
2

 ∑x  
×
2
∑ x = 146 + 160 × 30 = 4946,∑ x = 30  var(x) +  30   = 815 744


×
80
( )
( ) ⇒ standard deviation = $3.23
1024 − 146
30
30
b
Cumulative frequency
×
50
3
130 140 150 160 170 180 190 200 210 220
×
Median ≈ $270
12 a
Mass, m
(grams)
30 
m<
40
40 
m<
50
50 
m<
60
60 
m<
80
80 
m<
120
∑ fx = 30 660,∑ fx 2 = 10 984900
Frequency
13
37
56
8
6
30660
= $307 (3 s.f.)
100
Standard deviation ≈
Cumulative
frequency
13
50
106
114
120
100 200 300 400 500 600 700 800 900 1000
Monthly shopping bill ($)
c
×
80
1
9
×
100
The upper quartile is the 185th result, which
also lies in the 170–180 class.
Mean ≈
10 984 900
− 306.6 2 = $126 (3 s.f.)
100
10
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1
WORKED SOLUTIONS
So here, the standard deviation is more than
three-quarters of the IQR.
90
Note: if the distribution was more
symmetrical and less skewed, the rule would
work better.
Cumulative frequency
120
13 a Mean = 4930 = 290 kg
17
290 + 2(10.5) = 311 kg
b New mean = 4930 + 292 = 290.1 kg
18
2.5
×
50
14 a Width =
= 12.5 cm
10
b Height = 4 × 34 = 9.7 cm
14
60
30
c
90
120
90
Mass, m (g)
i Median ≈ 51g
b
ii LQ ≈ 44 g, UQ ≈ 57 g
Mass, m
(grams)
30 
m<
40
40 
m<
50
50 
m<
60
60 
m<
80
80 
m<
120
Frequency
13
37
56
8
6
Class width
10
10
10
20
40
Frequency
density
1.3
3.7
5.6
0.4
0.15
Cumulative frequency
60
30
Q3 ≈ 38
80
×
70
60
×
50
40
30
Q1 ≈ 21
20
×
10
Median ≈ 29
×
0
20
40
60
80
100
Time (milliseconds)
6
52nd value = median ≈ 29
5
Frequency density
×
×
100
26th value = Q1 ≈ 21
78th value = Q3 ≈ 38
4
IQR ≈ 17
d Median is best due to the extreme values in
the data set.
15 a The given bar has an area of 9 cm2, so each plant
is represented by an area of 9 cm2. That means
29
3
2
1
60
30
90
Mass, m (g)
c For this data, IQR ≈13 g.
∑ fm ≈ 53
Mean ≈
120
s.d. ≈
∑ fm 2 − 532 = 13.7 (3 s.f.)
120
120
the area of the 25  m < 35 bar is 9 × 4 cm2.
29
The width of the 25  m < 35 bar must be double
the width of the 5  m < 10 bar, at 3 cm.
That means the height of the 25  m < 35
bar is 9 × 4 = 0.414 cm (3 s.f.).
3 × 29
b Mean = ∑
fx
= 10.7 kg
∑f
Standard deviation =
∑ fx 2 − x 2 = 6.84 kg
∑f
11
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1 Representation of data
16 a Mean = ∑
fx
=
∑f
18 a Mean =
2
23 × 105 + 33 × 64 + 43 × 42 + 53 × 30 + 64 × 21 + 80 × 10
272
Variance =
= 37.011...
∑ fx 2 − x 2 =
∑f
(
437185 10 067
−
272
272
)
b ∑(t − 2500) = −2033, ∑(t − 2500) = 1 986 329
2
2
19 a Australia
853
98210
976641
885430
8611
542
= 15.4 (3 s.f.)
= 15 years 5 months (nearest month)
b The modal class is the one with the
highest frequency density. This is the class
containing 18–27 year olds.
c
773 071 329  310 467 
−
= 15 626.11...
125
 125 
s.d. = 125 s (2 minutes 5 seconds)
= 37 years, no months
s.d. =
310 467
= 2484 s (41 minutes 24 seconds)
125
48
49
50
51
52
53
England
0148
233679
1147888
246699
034
Key: 0 48 3 =
6
48.0 seconds Australia,
48.3 seconds England
Age 18–27 28–37 38–47 48–57 58–69 70–89
Freq
105
64
42
30
21
10
CF
105
169
211
241
262
272
b
Australia
England
300
48.0
50.0
51.0
52.0
53.0
54.0
Time (seconds)
250
Cumulative frequency
49.0
cBoth diagrams show the data well. Box -andwhisker plots show that overall the English
cricketers have slightly quicker times. Although
Australia has a smaller range, the interquartile
range of the two teams is similar. The stem-andleaf diagram has the advantage that the original
data is preserved.
200
150
100
50
20 For distribution 1, (lowest value, Q1, Q2, Q3,
highest value) = (8, 10, 14, 22, 30)
0
For distribution 2, (lowest value, Q1, Q2, Q3,
highest value) = (8, 14, 22, 26, 30)
15
30
45
60
75
90
Age (years)
edian ≈ 33, LQ ≈ 24, UQ ≈ 46, IQR ≈ 22 (all
M
in years)
d The modal class would be the most meaningful
measure to choose here, although it would help
to know what percentage of all drivers are in
each class to make a fully-informed choice.
17 IQR = 6
∑ x = 55 + a,∑ x 2 = 385 + a 2 ⇒
0
5
10 15 20 25 30
The values for distribution 2 are significantly
higher, as shown by the differing values for the
median. The measures of spread, however, are the
same here; the variation in each set of data would
seem to be the same.
2
2
s.d. = 385 + a −  55 + a  = 6
 11 
11
This gives 10a2 – 110a – 3146 = 0, and so a = 24
12
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WORKED SOLUTIONS
1
Mathematics in life and work
1
Drug Patients L
2
Q
L
×
Q
C (£
1000)
CE Continue?
A
15
1
0.9
0.9
10
11.1
no
B
31
2
0.7
1.4
12
8.6
no
C
64
3
0.7
2.1
15
7.1
yes
D
42
4
0.6
2.4
8
3.3
yes
E
34
5
0.4
2.0
17
8.5
no
F
17
6
0.3
1.8
3
1.7
yes
∑ fx = 2447,∑ fx 2 = 33 031
Mean = 2447 = 12.1 (3 s.f.)
203
Standard deviation =
2
33 031  2447 
−
= 4.17 (3 s.f.)
 203 
203
3
Median = 102nd value = 3 years
Q1 = 51st value = 3 years
Q3 = 153rd value = 5 years
4
CE is the largest for drug A.
5
NICE would wish to discontinue drugs A, B and E,
and continue with drugs C, D and F .
13
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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS
2 Probability, permutations and combinations
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication
of the answer.
Prerequisite knowledge
1
b P(both yellow) = 16 = 0.16
100
c P(one blue and one yellow) = 0.24 + 0.24 = 0.48
a
x
Exercise 2.1A
K
B
1
59
24
30
2
3
?
Alternatively: There are 2! ways to arrange
the two S books. There are 6! ways to
arrange the remaining 6 books. Finally there
are 7 possible positions for the S books in
relation to the N books, so 7 × 6! × 2! = 10 080
ways in total.
a 1
3
b 4
15
3
c 11
15
a P(K and J) = 0.2
So the number of ways where the two S
books are not together is 40 320 − 10 080 =
30 240.
b P(K and H) = 0
c P(H′) = 1 − 0.35 = 0.65
 2  6   2  6 
c     +     = 55
 1   3   2  2
d P(K′ and J′ and H′)
= 1 − (0.25 + 0.2 + 0.1 + 0.35) = 0.1
4
4
e P(K or H) = 0.25 + 0.2 + 0.35 = 0.8
a
6
10
4
10
Blue
4
10
24
Yellow
100
24
Blue
100
Blue
6
10
The group must consist of two from one nation
and one from each of the others.
So the total number of possible choices is:
 5
 4
 3
 2 × 4 × 3 × 4 + 2 ×  2 × 5 × 3 × 4 +  2 × 5 × 4 × 4
36
100
6
10
= 1440
Yellow
4
10
14
b 8 × 7 × 6 = 336
5! × 21! = 6.13 × 1021 (5! arrangements for the
vowels, and then a following 21! arrangements for
the remaining letters.)
a 8! = 40 320
b The number of ways where the two S books
are together = 7! + 7! = 10 080.
b 120 − 59 − 30 − 24 = 7 members
2
a 8! = 40 320
Yellow 16
100
5
a The number of ways of arranging all 12 books
is 12! = 479 001 600.
The number of ways of arranging the eight Plato
books is 8! = 40 320, while the number of ways of
arranging the four Socrates books is 4! = 24.
So the probability of having Plato-Socrates is
8!4! while the probability of Socrates-Plato
12!
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WORKED SOLUTIONS
2
m(m − 1) n(n − 1) = 9 n(n − 1)(n − 2) m ⇒ (m − )
is 4!8! , and so the probability of having all
5 − 2) 1
m
) n(n − 1) 6= 9 n(n − 1)(n
12!
2
2(m − 110
m ⇒ 5(m − 1)
2
2
10
6
− 2)
the Plato books andmall
(mthe
− 1)Socrates
n(n − 1) =books
9 n(n − 1)(n − 2) m =⇒3(5n(m
− 1) = 3(n − 2)
2
2
10
6
together is 2 × 8!4! = 2 .
n − 2)
12! = 3(495
So the smallest values are m – 1 = 3, n – 2 = 5, or
b We now need the probability of Socratesm = 4, n = 7.
Plato-Socrates, where the arrangement is
10 a 3 × 8! = 120 960 b 3 × 3 × 6! = 6480
0–8–4 or 1–8–3 or 2–8–2 or 3–8–1 or 4–8–0.
c 5 × 4 × 6! = 14 400
The probabilities here are:
8! 4! + 4 × 8! × 3! + 4 × 3 × 8! × 2!
12!
12!
12!
×
×
×
!
×
1
!
4
!
×
8!
4
3
2
8
+
+
12!
12!
Exercise 2.2A
1
= 5 × 8! × 4! = 1
12!
99
6
7
a 6! = 60
2!3!
b 5! × 2 = 20
3!2!
c 4! = 4
3!
Probability of ‘Aba’ house = 0.26.
robability of not ‘Qom’ house or ‘Aba’
P
house is therefore 1 − (0.31 + 0.26) = 0.43.
2
Probability of selecting a square ticket
10
number =
= 0.1.
100
Therefore the probability of not selecting a
square ticket number = 1 − 0.1 = 0.9.
In these, the Ls can be arranged in 2!
ways, as can the Es and the Ss. Thus the
total number of distinct arrangements is
8!
= 7! = 5040.
2!2!2!
b Count instead the number of arrangements
where the two Ls DO appear together.
3
So the number of distinct arrangements
where the two Ls do not appear together is
aK(H)Q(H) appears in 7! arrangements, while
Q(H)K(H) appears in another 7!, so there are
2 × 7! ways for this to happen altogether
= 10 080 arrangements.
4
9
 m  n
 m  n
You have     = 0.9    
 1   3
 2   2
a
6! = 3
( ) 28
8!
3!
b P(green and yellow together)
7! × 2 = 1 .
=
3! 8 !
4
3!
3
P(not together) = .
4
6!
3
c 8! =
28
3!
b K(S) and K(C ) are together in 10 080
arrangements, so they are not together in
8! – 10 080 arrangements = 30 240
arrangements.
c If K(H) and Q(H) are together and K(S) and
K(C ) are together, then there are (regarding
K(H)Q(H) as one card and K(S)K(C) as
another) 6! × 2! × 2! = 2880 arrangements.
c Ten numbers end in 5, ten start with 5, and
one number does both, so 19 numbers
feature the digit 5. The probability is 0.19.
20 = 3x + 8
12 = 3x
x=4
4
Probability of selecting a black ball =
= 0.2.
20
Therefore, the probability of not selecting a
black ball = 0.8.
5040 – 1260 = 3780.
8
a Half of the tickets are even numbers so 1 or 0.5.
2
b Square ticket numbers are 1, 4, 9, 16, 25, 36, 49, 64,
81, 100. There are 10 square ticket numbers.
a
Regarding the Es and Ls and Ss as distinct
letters, there are 8! arrangements of the letters.
This is (regarding LL as one letter)
7!
= 1260.
2!2!
a
If there are 200 students in total, 46 must be in
Nis. The probability is therefore 46 = 0.23.
200
b Probability of ‘Qom’ house = 0.31.
( )
5
 10
 
a  4 = 2
 30 261
 4 
 n
n!
Using   = r !(n − r )!
r
15
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2 Probability, permutations and combinations
Thus P(team of five contains the goalkeeper,
 20
 4 
= 323
b
 30 1827
 4 
c
7
9
 10
 13
 7
 2  × 13 × 7 +  2  × 10 × 7 +  2 × 10 × 13
12 285 13
=
27405 29
Heads
Tails
Spades
SH
ST
Hearts
♥H
♥T
Diamonds
DH
DT
Clubs
CH
CT
b Red suits = hearts and diamonds, so the
probability is 2 = 1
8 4
c
H, H
d
H, T
T, H
3
4
b P((H or T )) = 1 – P(H or T )
10 a P(H) =
= 1 – P (spade/diamond/club or ten)
= 1 –  3 + 1  = 3
 4 52  13
c P((B or (T or H))) = 1 – P(B or (T or H ))
= 1 – P(B or T or H )
=1
– P(T or H ) (since if B happens, H and T
happen)
 3 1
=1– + 
 52 4 
9
=
13
Exercise 2.3A
All eight outcomes are equally likely.
8
 10  22
 2   2  × 2
20790 15
=
=
= 0.163 (3 s.f.)
c
127512 92
 24
×
4
×
3
 4 
Probability of Alan getting a job (up to age 24)
≈ 1 − 0.132 = 0.868.
The above figure is a crude estimate. Is Alan
better or worse qualified than the average
unemployed Australian? Does he spend more
or less time than the average seeking work? Are
the job opportunities for Australians the same
as for those from overseas? Is Alan willing to
compromise on levels of pay? There are a large
number of unknowns. It could be that since
Alan is choosing to move to Australia, he may
have specific work in mind that increases his
chances of finding a job.
a
 24
a   × 4 × 3 = 127 512
 4
 14  13  10
 1   2   1 
10 920
65
b
=
=
= 0.0856 (3 s.f.)
127 512 759
 24
4
3
×
×
 4 
 30
 4 
=
6
two defenders and two attackers) = 100 = 50 .
462 231
1
x
0.3
T, T
Spades
SHH SHT STH STT
Hearts
♥HH ♥HT ♥TH ♥TT
Diamonds
DHH DHT DTH DTT
Clubs
CHH CHT CTH CTT
B
A
0.2
0.1
0.4
All 16 outcomes are equally likely.
a 0.6
2 1
=
16 8
b 0.1
Total number of choices for the team of five is
11
C 5 = 462 .
Total number of teams with the goalkeeper, two
defenders and two attackers = 5C 2 × 5C 2 = 100 .
16
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2
WORKED SOLUTIONS
2
x
5
x
Pianists
B
A
42
guitar
violin
0.15
0.15
0.35
36
10
68
0.35
a 0.7
P(piano but not violin and not guitar) = 42 = 7
156 26
b 0.5
c 0.65
3
6
x
x
L
J
S
K
0.15
x
y
z
0.25
0.25
0.35
w
P( J or L but not both) = 0.4.
x + y = y + z = 3y, x + y + z = 0.75,
so 5y = 0.75, y = 0.15, x = z = 0.3
7
x
Gardeners
a 0.15
b 0.45
hoe
c 0.55
d 0.25
x
e 0.3
4
fork
x
10
7
24
x
14
x
9
spade
B
A
3x + 64 = 100, so x = 12.
0.1
0.4
0.2
P(spade) = 14 + 12 + 7 + 12 = 0.45
100
8
0.3
a 0.4
b 0.1
The probabilities must add up to 1, so 4x + 4y = 1,
x + y = 0.25.
P(B) = 0.3 implies 3y + x = 0.3.
Subtracting the two equations,
2y = 0.05, so y = 0.025, x = 0.225
Thus P(C) = 3x + y = 0.7.
c 0.2
17
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2 Probability, permutations and combinations
9 Say the number of members doing all three
Saystrokes
the number
is a. of members doing all three strokes is a.
x
crawl
Exercise 2.4A
1
x
B
A
butterfly
30–(39–a)
24–a
33–(43–a)
a
15–a
19–a
30–(34–a)
1
4
1
5
?
11
20
backstroke
So 7 = 3a–23 and so a = 10
a 0
1
b 5
x
crawl
butterfly
1
14
10
5
11
c 20
0
2
9
6
5
backstroke
a P (all three strokes) = 10 = 1
50 5
5
b P (none of these) =
=1
50 10
3
10 x
A
B
ar
ar
3
ar 2
Y
a 5
8
a + ar + ar2 + ar3 = 1 ⇒ a =
8
15
B
3
8
Y
c 4×3=1
9 8 6
d 4×5+5×1=5
9 8 9 2 9
Wears a
scarf
Does not
wear a scarf
Total
Wears
a hat
5
12
1
4
2
3
Does not
wear a hat
1
12
1
4
1
3
Total
1
2
1
2
1
1
P ( S ∩ H ′ ) 12 1
=
=
b
4
1
P ( H ′)
3
5
8
b 1
2
Thus P(A) = 1 and P(B) = 1
3
5
18
1
Let A be ‘the sum on the dice is 4’, so P(A) = .
12
Let B be ‘the scores are equal’, so P(B) = 1 .
6
8
2
P(A ∪ B) =
= , which does not equal 1 + 1 = 1 .
36 9
12 6 4
Thus A and B are not mutually exclusive.
1
B
2
B
5
1
9
Y
2
4
9
a
11 a
A and B are mutually exclusive if and only if
P(A ∪ B) = P(A) + P(B).
4
a
R
4
7
2
7
B
1
7
G
2
3
H
8
21
1
3
T
4
21
1
2
H
1
7
1
2
1
2
1
2
T
H
T
1
7
1
14
1
14
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WORKED SOLUTIONS
b P(red and heads) =
8
21
2 7 1 4
4
a 7 × 13 + 2 × 13 = 13
1
7
b P(at least one blue)=
1 – (P(RR) + P(GG) + P(RG) + P(GR))
P(blue and heads) =
P(green and heads) =
P(heads) =
1
14
8 1 1
25
+ +
=
= 0.595
21 7 14 42
5
0.45
L
0.55
S
B
0.6
= 1−
W
0.6
B
0.1
W
0.9
0.5
B
W
0.5
B
0.2
W
FR
0.75
2 3 1 6
4
36
×
− ×
−
=
7 13 2 13 13 91
c P(both the same) = P(RR) + P(GG) + P(BB)
2 3 1 6
3
2 30
= 7 × 13 + 2 × 13 + 14 × 13 = 91
P (different) = 1 – 30 = 61
91 91
8
a
x
bus
L
0.25
0.4
0.4
train
45
6
18
S
0.8
B
a 0.6 × 0.45 × 0.4 + 0.4 × 0.25 × 0.5 = 0.158
75
6
= 0.0417
144
24
51
P(B) × P(T ) =
×
= 0.0590
144
144
b 0.6 × 0.45 × 0.6 + 0.6 × 0.55 × 0.9 + 0.4 ×
0.25 × 0.5 + 0.4 × 0.75 × 0.8 = 0.749
P(B ∩ T ) =
c 0.4 × 0.25 × 0.5 + 0.4 × 0.75 × 0.8 = 0.29
6
a i 0.65 × 0.4 × 0.25 = 0.065
( ) ( )
These are not equal, so B and T are not
independent.
ii 0.2 × 0.85 × 0.75 = 0.128
iii0.1 × 0.45 × 0.55 + 0.1 × 0.45 × 0.45 + 0.1 ×
0.55 × 0.55 + 0.45 × 0.55 × 0.9 = 0.298 (3 s.f.)
b
x
b iP(both walk) = 0.9 × 0.25
P(both cycle) = 0.55 × 0.7
bus
train
P(same method ) = 0.61
45
6+x
18 – x
ii P(different methods) = 0.39
2
13
7
B
4
13
3
14
7
14
3
13
G
4
13
4
14
2
3
13
R
3
13
B
75
7
13 G
P(B ∩ T ) = P(B) × P(T ),so
( )
(6 + x) (51 + x)
24
=
×
,
144
144
144
R
which solves to give x = 3
B
6
13 G
9
a A and B are mutually exclusive.
b
x
C
R
B
7
13 G
A
3
6
2
4
1
5
B
R
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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS
Probabilities of 1, 2, 3, 4, 5, 6 are kp, p, p, p, p, p
so p(5 + k) = 1, and p = 1 .
5+k
P(B)P(C) = P(B∩C)⇒
k+2 3
2
=
⇒k=4
5+k5+k k+5
5
x
F
E
1
10
10 Since A and B are mutually exclusive, P(A∩B) = 0.
If they are also independent, then P(A∩B) = P(A)P(B).
Thus P(A)P(B) = 0, and so either P(A) = 0 or P(B) = 0
or both.
P(E ∩ F )
P(F )
= 0.15 ÷ 0.5
a P(E | F ) =
P(spade) = 1 , P(jack) = 1 , P(spade and jack) = 1
52
4
13
1
2
= 0.3
P(jack | spade) = 52
= 1
1
13
4
Or, since the events spade and jack are clearly
independent, P(jack | spade) = P(jack).
6
P(W) = 14 , P(W ∩ H) =
31
31
P (H ∩ W )
P (W )
6 14
=
÷
31 31
3
=
7
1
1
1
P(H, T or T, H) = , P(H, H) = , P(T, T) =
2
4
4
b P(E′ ∩ F ) = P(F ) − P(E ∩ F )
= 0.5 − 0.15
= 0.35
c P(E′ ∩ F′) = 1 − P(E ∪ F )
= 1 − 0.6
P(H | W) =
3
1
P(T , T )
1
4
a P(T, T | T) = P(at least one T ) = 3 = 3
4
6
a
5
6
Red
Yellow
5
11
M
0.1
0.4
P(M P)
P(P)
= 0.4 ÷ 0.5
= 0.8
P(M ∩ P ') 0.2 2
d P(M | P′) =
=
=
0.5 5
P(P ')
c P(M | P) =
Yellow
Red
Red
Yellow
2
6
4
6
2
6
Yellow
Red
3
6
3
6
0.2
Yellow
Red
Yellow
 1 5 5  1 6 4
b P(B) =  × ×  +  × × 
 2 11 6   2 11 6 
+  1 × 6 × 4  +  1 × 5 × 3 
 2 11 6   2 11 6 
0.3
a P(M ∪ P) = 0.7
2
b P(P | M) = P(P ∩ M ) = 0.4 =
0.6 3
P(M )
6
11
Red
1
6 4
6
Yellow
6
11
x
P
Red
5
11
6
12
1
= 32 = 2
3
4
4
= 0.4
6
12
b P(H, T | T) = P(H ,T or T , H )
P(at least one T )
7
20
2
5
Exercise 2.5A
1
3
20
=
2
3
c P(A ∩ B) = P(R, R, R) + P(Y, Y, R)
= 25 + 5
132 44
=
10
33
20
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WORKED SOLUTIONS
d (R, Y, Y) and (Y, R, Y) are the only disallowed
combinations.
So P(A ∪ B) = 1 – 1 − 1
11 11
=
b P(G) = 0.1 × 0.3 + 0.9 × 0.5 = 0.48
9
9
11
25
a
Red
7
9
Blue
)(
=
1
1
10 a 4 = P(R | S) + P(S | R)
P(S)
P(R)
+
P(R ∩ S) P(S ∩ R)
P(S) + P(R)
=
P(R ∩ S)
0.8
=
P(R ∩ S)
=
Heads
4
5
(
b 0
e P( A | B ′ ) =
Tails
2
5
1
5
2 2
7 4
b P(tails) = 9 × 5 + 9 × 5
a 1
Heads
3
5
2
9
P(W ∩ G ) 0.03
=
= 0.0625
P(G )
0.48
P( A ∩ B ′ ) P( A ∩ B ′ )
=
P(B ′ )
1 − P(B)
P( A) − P( A ∩ B)
=
1
3
= 3(0.5 − 0.2) = 0.9
5
( 33 ) 8
7
c P(W | G ) =
c P(A ∩ B) = P(B | A)P(A) = 0.2
P( A ∩ B) 0.2 2
d P(B) = P( A | B) = 0.3 = 3
25
5 10
e P(R, R, R) + P(Y, Y, Y) = 132 + 44 = 33
=
P(all red | same colour) = 132
10
Tails
)
Thus P(R ∩ S) = 0.2.
2
5
1
8
b 4=
+
=
⇒ P(S | R) =
3
3P(S | R) P(S | R) 3P(S | R)
32
45
c P(heads) = 13
45
P(R ) =
P(blue | heads) =
P(blue ∩ heads)
P(heads)
=
7 13
÷
45 45
=
7
13
1
(
)(
a
0.3
0.1
)
( 3060 ) = 30 ÷ 48 = 0.625
48
( 160
)
 3   4   5  3   4   5 
a   ×  ×  +  ×  × 
 2  2   3  2  3   2
 3  4   5 
+   ×   ×   = 360
 3  2   2
G
b 10! × 3! = 2177 2800
W
0.7
0.5
0.9
a 1 − 56 = 0.65
160
b 24 + 32 = 0.35
160
c
2
53
=
81
P(R ∩ S) 0.2
=
= 0.3, so P(S) = 0.5
2
P(S | R)
3
Exam-style questions
d Probability of same colour = P(red and red) +
P(blue and blue)
2 2
7 7
=
×
+
×
9 9
9 9
8
2
G'
G
3
a P(Uche does not) = 1 − 0.62 = 0.38
P(Eli does not) = 1 − 0.17 = 0.83
P(Ellis does not) = 1 − 0.68 = 0.32
0.38 × 0.83 × 0.32 = 0.101 (3 s.f.)
W'
b P(at least one) = 1 – P(none)
0.5
G'
= 1 – 0.100 928 = 0.899 (3 s.f.)
21
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2 PROBABILITY, PERMUTATIONS AND COMBINATIONS
c P(two hit) = 0.62 × 0.17 × 0.32 + 0.62 × 0.83 ×
0.68 + 0.38 × 0.17 × 0.68
= 0.428 (3 s.f.)
0.428
P(two hit | at least one hits) =
= 0.476 (3 s.f.)
0.899
4
6
a
1
3
R
x
a
1
3
climbing
26
2
4
35
1
4
40
35
11
19
16
1
3
1
4
2
3
b 35 out of 185 do none of the activities.
G
Therefore the probability of a student
doing none of the activities is 35 ÷ 185
= 0.189 (3 s.f.).
Therefore the probability of a student doing
bowling only is 19 ÷ 185 = 0.103 (3 s.f.).
( )
( )
R
a
Time taken
 1 min
Time taken
> 1 min
Total
Question
right
1
8
13
24
2
3
Question
wrong
1
8
5
24
1
3
Total
1
4
3
4
1
2x
2
3
B
1
3
x
2
3
G
13
()
1
3
1
3
1–3x
= 24
= 13
2
16
3
1
3
1
3
35
185
104
185
P(slowly and right)
P(right)
1
6
1
6
G
1
12
R
1
6
B
1
12
c
= 0.336 (3 s.f.)
b P(slowly | right ) =
1
6
1
1 1
b P(B ∩ G) = P(B G) + P(G B) = 12 + 12 = 6
c 19 out of 185 do bowling only.
d P(no activities | not bowling) =
B
R
B
table tennis
1
6
G
2
3
3
bowling
5
1
3
R
1
3
R
1–3x
3
B
1–3x
3
1–3x
3
R 4x
3
G
G
2x
3
R
2x
3
B
x
3
(1 − 3x) + 2x = 1 ⇒ 5 − 15x + 10x = 3
3
3 5
2
So x =
5
22
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WORKED SOLUTIONS
7
a p = 15, q = 9, r = 7, s = 1
b P(art | music ) = P(both art and music)
P(music)
=
9
32
16
32
= 9
( ) 16
c P(art) = 24 = 3 ≠ 9 , so these are not
32 4 16
independent events.
d
8
7 15
×
= 0.106 (3 s.f.)
32 31
 6   3  2  6   3   2  6   3  2
a       +       +       = 144
 2  1   1   1   2  1   1   1   2
b 2 × 9! = 725 760
c 6 × 5 × 7! = 151 200
9
a 47 + 62 – 100 = 9 boys both cycle and swim.
91 = 0.91 = P (student does just one sport).
b
100
P(cycling only)
c P(cycling | one sport only) =
P(one sport only)
= 0.38 ÷ 0.91
= 0.418 (3 s.f.)
d Because some boys do both sports.
Alternative answer: P(C ∪ S) not equal to
P(C) + P(S)
11 The game will be fair if the probability of spinner 1
beating spinner 2 is equal to spinner 2 beating
spinner 1. The same applies between spinner 1 and
spinner 3, and between spinner 2 and spinner 3.
Probability of spinner 1 beating spinner 2:
=
( 63 × 15 ) + ( 62 × 35 ) + ( 16 × 45 ) = 1330
Probability of spinner 2 beating spinner 1:
=
( 15 × 63 ) + ( 15 × 63 ) + ( 15 × 65 ) + ( 15 × 1) = 1730
in 1 versus 3, P(1 win) = 16 , P(3 win) = 14 ,
30
30
14
in 2 versus 3 P(2 win) =
25
These probabilities are unequal; therefore the
game is unfair.
12 a
0.02
Misshapen (M)
0.98
Not misshapen (M′)
0.03
Misshapen (M)
0.97
Not misshapen (M′)
0.05
Misshapen (M)
0.95
Not misshapen (M′)
J
0.25
0.45
K
0.3
L
e P(C & S) = 9 ÷ 100 = 0.09 ≠ 0.47 × 0.62
10 a Treat the jacks as a block, so you have the
queens, the kings and the jack block, which is
9 things to arrange. There are 4! ways to arrange
the jacks within the jack block.
b P(M′ ∩ J) = 0.25 × 0.98
Total number of arrangements is 12!, so
9! 4! = 0.0182 (3 s.f.).
P(4 jacks together) =
12!
b Ways to arrange the four kings = 4!, and the
other cards = 8!
= 0.0335
So P(2K at one end, 2K at the other)
4! × 8!
= 12! = 0.00202 (3 s.f.)
c 3 ×  4  4  4 = 288
 2  1   1 
d
288 = 0.582 (3 s.f.)
 12
 4 
2
= 0.245
c P(M) = P(M ∩ J ) + P(M ∩ K) + P(M ∩ L)
= 0.005 + 0.0135 + 0.015
P(K '∩ M )
P(M )
[
P
(
M ∩ J ) + P(M ∩ L)]
=
P(M )
d P(K ′ | M) =
= 0.02 ÷ 0.0335
= 0.597 (3 s.f.)
5
of the numbers from 2 to 10 are even, so the
13 a
9
answer is 5 .
9
b Two odd numbers add to give an even
number, but multiply to give an odd number,
so S and T are not mutually exclusive.
c P(S) = P(two cards are both even)
+ P(two cards are both odd)
= 20 × 19 + 16 × 15 = 31
36 35 36 35 63
23
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2 Probability, permutations and combinations
17 a
P(T) = P(two cards are both odd)
16 15
4
×
=
=
36 35 21
T
0.3
T'
0.75
T
0.25
T'
0.25
T
0.75
T'
S
4
4 31
≠
× , so S and T are
21 21 63
not independent.
P(S ∩ T) = P(T) =
0.35
You could argue instead that P(S | T) = 1 ≠ P(S).
14 a 9 × 8 × 7 × 6 × 5 = 15 120
0.4
b Digits that add up to 9 or 18. (1, 2, 6), (1, 3, 5),
(2, 3, 4), (9, 8, 1), (9, 7, 2), (9, 6, 3), (9, 5, 4),
(8, 7, 3), (8, 6, 4) are the possible triplets. Each
produce 6 numbers, so 54 in total.
54 = 3
9 × 8 × 7 = 504, so P(divisible by 9) =
504 28
c Each group has 3 members. Number of ways:
 3   3  3
=       × 3 = 81
 2  1   1 
P
0.25
F
b P(T ) = 0.35 × 0.7 + 0.4 × 0.75 + 0.25 × 0.25 = 0.608
d The first three cards he picks must add up to a
1
multiple of 3, so the answer is .
3
15 a 0.8 − 0.36 = 0.44
c P(F | T ) =
R = Red
G = Green
c (0.44 × 0.56) + (0.36 × 0.66) + (0.2 × 0.25)
= 0.534
d 1 − 0.534 = 0.466
(0.466)5 = 0.022 (3 d.p.)
16 Let us assume that there are 100 students at
the school (probabilities are not affected by
school size);
x
B
G
2
3
4
5
6
1
1
2
3
4
5
6
2
2
4
6
8
10
12
3
3
6
9
12
15
18
4
4
8
12
16
20
24
5
5
10
15
20
25
30
6
6
12
18
24
30
36
5
6
R = Red
1
1
21%
a 0.26
G = Green
22%
1
b
C
31%
P(F ∩ T )
0.252
=
= 0.103(3 s.f.)
P(T )
0.6075
18 a
b 1 − 0.44 = 0.56
26%
0.7
2
3
4
yes
yes
3
4
2
yes
yes
yes
5
6
yes
yes
yes
b 0.21
c P(B ∩ C ) = 0.26 = 0.456(3 s.f.)
P(C )
0.57
d P(G)P(C ) = 52 57 ≠ 31 = P(G ∩ C ) , so not
100 100 100
independent.
P(RG is square) = 8 = 2
36 9
5
P(even ∩ square) 36 5
=
=
P(square)
8 8
36
5
3
d P(odd | square) = 1 − =
8 8
c P(even | square) =
24
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WORKED SOLUTIONS
2
 5  24
19 a     = 2760
 3  2 
b
 5  24  5  24
 1  4  +  2  3  = 73 370
20 a 12! = 6 × 11!
2!
10!
b 8 × 4 × 2! = 16 × 10!
c 8 × 8!
d
2
1485
21 a
Red even
Red odd
Total
Green
even
1
3
1
6
1
2
Green
odd
1
4
1
4
1
2
Total
7
12
5
12
1
b P(product even | sum even) =
P(green even or red even | both odd or both
even)
P(both even)
=
P(both odd or both even)
=
(
1
4
3
=
7
1+1
3 4
)
Mathematics in life and work
1
Gearbox
Exhaust
Expected cost
0.9
Okay 0.855 $0
$0
0.95
0.1
Fail
$47.50
0.05
0.9
Okay 0.045 $3000
$135
0.1
Fail
$17.50
Okay
0.095 $500
Fail
2
3
0.005 $3500
$200
The premium should be $220 per year.
25
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3 Discrete random variables
3 Discrete random variables
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication
of the answer.
Prerequisite knowledge
1
2
a {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
3
b
8
1
2
4 1
=
c
8 2
3
1 1 1
+ =
6 6 3
2
a
3
1 ×1 ×2= 2
36
6 6
33
Mean =
= 1.1
30
Variance = 77 – 1.12 = 1.36 (3 s.f.)
30
4
b
4
5
6
8
a 6! = 6 × 5 × 4 … = 720
b 10C4 = 10! = 210
(6!4!)
9
10
Exercise 3.1A
1
a
12
a
15
x
2
3
4
5
6
7
8
9 10 11 12
P(X = x)
1 2 3 4 5 6 5 4 3 2 1
36 36 36 36 36 36 36 36 36 36 36
16
18
20
b
24
0.180
25
0.160
P (X = x)
0.140
30
0.120
0.100
36
0.080
0.060
3
0.040
0.020
0.000
2
3
4
5
6
7
8
9 10 11 12 x
The distribution is unimodal and symmetrical.
26
P(A = a)
1
36
1
18
1
18
1
12
1
18
1
9
1
18
1
36
1
18
1
9
1
18
1
36
1
18
1
18
1
18
1
36
1
18
1
36
a 0.1 + 0.2 + a + 0.25 + 0.3 = 1
a = 1 – (0.1 + 0.2 + 0.25 + 0.3)
= 0.15
So P(Y = 1) = 0.15
b P(Y  0) = P(Y = −1) + P(Y = 0)
= 0.1 + 0.2
= 0.3
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3
WORKED SOLUTIONS
c P(Y > 1) = P(Y = 2) + P(Y = 3)
or
= 0.25 + 0.3
P(X  1) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.55
4
a
x
2
3
4
5
6
7
8
P(X = x)
1
16
1
8
3
16
1
4
3
16
1
8
1
16
7
b Possible prime numbers are: 2, 3, 5, 7
Therefore P(X is a prime number) =
1 +1+1+1 = 9
or 0.5625
16 8 4 8 16
5
4
5
6
P(X = x)
1k
2k
3k
4k
5k
6k
P(X = x)
1
21
2
21
3 =1
21 7
4
21
5
21
6 =2
21 7
∑p = 1
k + 2k + 3k + 4k + 5k + 6k = 1
21k = 1
k = 1 (substitute into the table)
21
b P(2  X < 4) = P(X = 2) + P(X = 3)
8
27
2
4
9
b P(X  1) = 1 − P(X = 0)
=1− 8
27
19
=
27
2
()
1
3×
3
2
9
2
×
3
2
3
4
P(R = r)
0.1r
0.1r
0.1r
k
P(R = r)
0.1
0.2
0.3
k
7k = 0.35
k = 0.05
3
2
1
0.4 + 0.25 + 4k + 2k + k = 1
There is one way of having 3 blues: B,B,B
1
r
9
The probability of all the possible outcomes add
up to 1, therefore:
There are three ways of having 2 blues:
B,B,G
B,G,B
G,B,B
P(X = x)
5
21
The value of k = 1 – (0.1 + 0.2 + 0.3) = 1 – 0.6 = 0.4
There are three ways of having 1 blue:
G,G,B
G,B,G
B,G,G
()
=
8
a There is one way of having 0 blues: G,G,G
( 23 )
2 1
+
21 7
= 2 + 4 +2=4
21 21 7 7
2
d P(2  X < 11) = 3
P(X = x)
=
c P(even number) = P(X = 2) + P(X = 4) + P(X = 6)
= 1 − P(X = 2)
=1− 1
6
5
=
6
1
2
3× ×
3
3
a
3
c P(X  3) = 1 − P(X < 3)
3
19
27
2
=1+1+1
6 6 6
1
= or 0.5
2
0
=
1
a
The outcome for each value of x is equal over six
possible outcomes.
1
Therefore k = .
6
x
4 2 1
+ +
9 9 27
x
b P(X < 7) = P(X = 2) + P(X = 3) + P(X = 5)
6
=
()
1
3
3
1
27
The probability that at least two under graduates
are late is:
P(X  2) = 1 – P(X  1) = 1 – (0.4 + 0.25) = 0.35
10 a T
he probability of all the possible outcomes add
up to 1, therefore:
0 + 3r + 9r + 4r + 3r + r = 1
20r = 1
r = 0.05
27
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3 Discrete random variables
b
0
1
2
3
4
5
P(C2 = c)
0.25k
1.25k
2.25k
3.25k
3.5k
4.5k
P(C2 = c)
1
1 1
×
=
4 15 60
1
5
1
×
=
4 15 12
3
9
1
×
=
4 15 20
13
1 13
×
=
4 15 60
7
7
1
×
=
2 15 30
3
9
1
×
=
2 15 10
c
0.25k + 1.25k + 2.25k + 3.25k + 3.5k + 4.5k = 1
1
15k = 1 ⇒ k =
15
1
59
P(C 2 1) = 1 − P(C 2 = 0) = 1 −
=
60 60
Exercise 3.2A
1
2
4
a i 0.2 + 0.15 + 0.35 + a + 0.1 = 1
a = 1 − (0.2 + 0.15 + 0.35 + 0.1)
aE(X) = −2 × 0.13 − 1 × 0.27 + 0 × 0.1 + 1 × 0.18
+ 2 × 0.22 + 3 × 0.1
= 0.2
= 0.39
b E(Y) = 2 × 1 + 4 × 5 + 6 × 1 + 8 × 1
3
6
12
12
31
= 5.17 (3 s.f.)
=
6
iiE(X) = 0
× 0.2 + 1 × 0.15 + 2 × 0.35 + 3 × 0.2
+ 4 × 0.1
a E(Z) = 5 × 0.4 + 6 × 0.3 + 7 × 0.2 + 8 × 0.1
x2
0
1
4
9
16
P(X = x)
0.2
0.15
0.35
0.2
0.1
= 1.85
iii
=6
b
z2
25
36
49
64
P(Z = z)
0.4
0.3
0.2
0.1
E(X2) = 0 × 0.2 + 1 × 0.15 + 4 × 0.35 + 9 × 0.2
+ 16 × 0.1
= 4.95
b i k + 2k + 3k + k = 1
E(Z 2) = 25 × 0.4 + 36 × 0.3 + 49 × 0.2 + 64 × 0.1
7k = 1
k = 1 = 1.14 (3 s.f.)
7
1
3
1
2
ii E(Y ) = −2 × 7 + 0 × + 2 × 7 + 4 × 7
7
8
=
7
1
2
3
1
2
iii E(Y ) = 4 × 7 + 0 × 7 + 4 × 7 + 16 × 7
32
= 4.57 (3 s.f.)
=
7
= 37
c (E(Z))2 ≠ E(Z 2) since (E(Z))2 = 62 = 36 ≠ 37 = E(Z 2).
1
2
3
4
5
6
1
0
1
2
3
4
5
2
1
0
1
2
3
4
3
2
1
0
1
2
3
4
3
2
1
0
1
2
5
4
3
2
1
0
1
6
5
4
3
2
1
0
d
0
1
2
3
4
5
P(D = d)
1
6
5
18
2
9
1
6
1
9
1
18
3
5
a
x
4
7
= 0.14
50
22
= 0.44
50
18
= 0.36
50
1
50 = 0.02
5
2
50 = 0.04
1
a
2
3
1
5
2
1
+1×
+2× +3× 1 +4×
6
9
9
18
6
1
+5×
18
b E(D) = 0 ×
=
35
= 1.94 (3 s.f.)
18
P(X = x)
b E(X) = 1 × 0.14 + 2 × 0.44 + 3 × 0.36 + 4 × 0.02
+ 5 × 0.04
= 2.38
28
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3
WORKED SOLUTIONS
1
1
1
4
1
b E(X) = 1 × + 2 × + 3 × + 4 × + 5 ×
10
8
8
10
8
1
37
or 3.7
+6× =
8 10
c P(X > 2.38) = P(X = 3) + P(X = 4) + P(X = 5)
= 0.36 + 0.02 + 0.04
= 0.42
6
x
P(X = x)
0
0.73 = 0.343
1
3 × (0.72)(0.3) = 0.441
2
3 × (0.7)(0.32) = 0.189
3
0.33 = 0.027
c
= 0.9
a + 4b + 2b + a = 1
2a + 6b = 1
(1)
−1 × a + 0 × 4b + 1 × 2b + 2 × a = 5
12
−a + 2b + 2a =
a + 2b =
10 a
5
12
5
12
12a + 24b = 5
(2)
8a + 24b = 4
(1) × 4
4a = 1
(2) – 4(1)
a=
b
1
4
()
Substitute a into (1): 2 1 + 6b = 1
4
6b = 1 −
=
1
2
4
5
6
2x
2
4
6
8
10
12
P(X = x)
1
10
1
8
1
8
4
10
1
8
1
8
y
2
4
6
8
P(Y = y)
2
20
4
20
6
20
8
20
y
2
4
6
8
y–1
2–1
4–1
6–1
8–1
P(Y = y)
2
20
4
20
6
20
8
20
E (Y–1) = 2
1
×
2
4
6
+ 4 −1 ×
+ 6 −1 ×
20
20
20
8 1
=
20 5
aVar(X) = (−2)2 × 0.13 + (−1)2 × 0.27 + 02 × 0.1
+ 12 × 0.18 + 22 × 0.22 + 32 × 0.1 – 0.092
= 2.60
b Var(Y) = 22 ×
z
12
P(Z = z)
x
1−x
4x + 12(1 − x) = 7
4x + 12 − 12x = 7
5 = 8x
5
x = = 0.625
8
12
0.625
0.375
∑p = 1
k + 4k + 4 = 1
8
1
k=
10
=
2
4
z
P(Z = z)
−1
Exercise 3.3A
4
a
3
+ 8 −1 ×
1
÷6
2
1
=
12
9
2
1
2
b=
8
1
E(2X) = 2 × 1 + 4 × 1 + 6 × 1 + 8 × 4 + 10 × 1
10
8
8
10
8
1
+ 12 × = 7.4
8
Alternative method:
37
E(2X) = 2 E(X) = 10 × 2 = 7.4
E(X) = 0 × 0.343 + 1 × 0.441 + 2 × 0.189 + 3 × 0.027
7
x
1
1
1
5
31
+ 42 ×
+ 62 × 3 + 82 × 6 −  
12
12
 6
107
= 2.97
36
d
1
2
3
4
5
6
P(D = d )
1
6
1
6
1
6
1
6
1
6
1
6
E(D) = 1 ×
2
1
1
1
1
1
1
+2× +3×
+4× +5× +6×
6
6
6
6
6
6
= 3.5
1
1
1
1
1
+ 22 × +32 × + 42 × + 52 ×
6
6
6
6
6
1
+ 62 × – 3.52
6
Var(D) = 1
2×
=
35
= 2.92
12
29
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3 Discrete random variables
3
a
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
6
= 11.5%
E(B) = 10 × 0.55 + 50 × 0.25 + −30 × 0.2
= 12%
b Var(A) = 1
02 × 0.6 + 252 × 0.25 + (−5)2 × 0.15
– 11.52
s
2
3
4
5
6
7
8
P(S = s)
1
16
2
16
3
16
4
16
3
16
2
16
1
16
= 87.75
Var(B) = 1
02 × 0.55 + 502 × 0.25
+ (−30)2 × 0.2 − 122
= 716
1
2
3
4
b E(S) = 2
× 16 + 3 × 16 + 4 × 16 + 5 × 16
c The expectation of stock B is higher, so on
average, he would get better returns on
his investment. The variance of B is much
higher so there is more risk. The stockbroker
may make a higher profit in stock B but
could also lose a lot more than in stock A.
3
2
1
+ 6 × 16 + 7 × 16 + 8 × 16
=5
1
2
3
4
Var(S) = 22 × 16 + 32 × 16 + 42 × 16 + 52 × 16
3
2
1
+ 62 × 16 + 72 × 16 + 82 × 16 − 52
4
a E(A) = 10 × 0.6 + 25 × 0.25 + −5 × 0.15
= 2.5
a 0.1 + p + 0.2 + 0.35 = 1
7
p = 1 − (0.1 + 0.2 + 0.35)
aP(R = 3) means having two rolls that are not 1
and then a third roll that is a 1, or 3 rolls that are
not 1.
2
=  5  × 1 +  5 
 6
6  6
= 0.35
b 2 × 0.1 + 5 × 0.35 + a × 0.2 + 8 × 0.35 = 5.95
=
4.75 + 0.2a = 5.95
0.2a = 5.95 − 4.75
b
= 1.2
a=
1.2
0.2
=6
5
r
1
2
3
P(R = r)
1
6
5× 1= 5
6 6 36
25
36
=
= 3.35
a a + 3b + 2a + b = 1
3a + 4b = 1
25
36
c E(R) = 1 ×
c Var(X) = 2
2 × 0.1 + 52 × 0.35 + 62 × 0.2
+ 82 × 0.35 − 5.952
1
5
25
+2×
+3×
6
36
36
91
= 2.53
36
( )
1
5
25
91
+ 32 ×
–
+ 22 ×
6
36
36
36
= 0.583
Var(R) = 12 ×
(1)
1 × a + 3 × 3b + 5 × 2a + 7 × b = 3.8
a + 9b + 10a + 7b = 3.8
11a + 16b = 3.8
(2)
12a + 16b = 4
(1) × 4
8
P(G = 2) = 12 × 11 = 33
20 19 95
a = 0.2
Substitute into (1):
g
0
1
2
P(G = g )
14
95
48
95
33
95
3(0.2) + 4b = 1
4b = 0.4
b = 0.1
b Var(Y) = 1
2 × 0.2 + 32 × 0.3 + 52 × 0.4 + 72
× 0.1 – 3.82
2
a Let G be the number of girls who are appointed.
P(G = 0) = 8 × 7 = 14
20 19 95
P(G = 1) = 8 × 12 + 12 × 8 = 48
20 19 20 19 95
Subtracting:
= 3.36
3
b E(G) = 0 × 14 + 1 × 48 + 2 × 33 = 6 = 1.2
95
95 5
95
Rounding to the nearest integer, one girl is
expected to be appointed.
30
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3
WORKED SOLUTIONS
9
Let A be the profit from new car Model A:
0
a
3000
2500
–500
P(A = a)
0.5
0.3
0.2
P(H = 3) = 1 × 1 + 5 × 1 = 7
8 3 16 3 48
E(A) = 3000 × 0.5 + 2500 × 0.3 + (−500) × 0.2 = 2150
b
Let B be the profit from new car Model B:
b
3500
3000
–1000
P(B = b)
0.4
0.4
0.2
5
 5
P(H = 0) =    1   1  = 1
32
 0  2   2 
P(H = h)
h
( 21 + 81 + 321 ) × 31
( 21 + 83 + 325 ) × 31
( 83 + 165 ) × 31
( 81 + 165 ) × 31
0
E(B) = 3500 × 0.4 + 3000 × 0.4 + (−1000) × 0.2 = 2400
1
E(B) > E (A), so the finance director would support
new car Model B.
2
P(H = h)
10 aLet X be the number showing on the spinner:
7
32
11
32
11
48
7
48
x
1
3
5
3
P(X = x)
1
3
1
3
1
3
4
5 ×1
32 3
5
96
5
1 ×1
32 3
1
96
If the spinner shows '1':
P(H = 1) = 1
2
P(H = 0) = 1 − 1 = 1
2 2
7 + 1 × 11 + 2 × 11 + 3 × 7 + 4 × 5
48
48
96
32
32
1
3
+5×
= or 1.5
96 2
c E(H) = 0 ×
If the spinner shows '3':
1
2
 3
P(H = 1) =    1   1  = 3
8
 1  2   2 
P(H = 2) =
 3  1  2  1  1 3
 2  2   2  = 8
Var (H) =  0 2 × 7 + 12 × 11 + 22 × 11 + 32 × 7 + 4 2 × 5 + 52 × 1  −

32
32
48
48
96
96 
 0 2 × 7 + 12 × 11 + 22 × 11 + 32 × 7 + 4 2 × 5 + 52 × 1  −  3 

32
32
48
48
96
96   2 
3
0
 3
P(H = 3) =    1   1  = 1
8
 3  2   2 
P(H = 0) = 1 − 3 − 3 − 1 = 1
8 8 8 8
If the spinner shows '5':
1
P(H = 2) =
 5  1  2  1  3 5
 2  2   2  = 16
3
Exercise 3.4A
1
a
4
1
 5
P(H = 4) =    1   1  = 5




2
2
32
 4
0
 5
P(H = 5) =    1   1  = 1
32
 5  2   2 
i P(X = 2) = (1 − 0.2) × 0.2 = 0.16
ii P(X = 5) = (1 − 0.2)4 × 0.2 = 0.081 92
iii P(X = 10) = (1 − 0.2)9 × 0.2 = 0.027 (3 d.p.)
b i P(X = 2) = (1 − 0.8) × 0.8 = 0.16
ii P(X = 5) = (1 − 0.8)4 × 0.8 = 0.001 28
iii P(X < 3) = P(X = 1) + P(X = 2)
= 0.8 + (1 − 0.8) × 0.8
= 0.8 + 0.16 = 0.96
2
 5
P(H = 3) =    1   1  = 5
16
 3  2   2 
5
= 15 or 1.42
12
4
 5
P(H = 1) =    1   1  = 5
32
 1  2   2 
2
c
i P(X = 4) =
( 78 ) ( 18 ) = 0.084 (3 d.p.)
3
ii P(X  4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
=
( )( ) ( ) ( 18 ) + ( 78 ) ( 18 )
7 1
7
1
+ 8 8 + 8
8
2
3
= 0.414 (3 d.p.)
iii P(3  X < 7) = P(X = 3) + P(X = 4) + P(X = 5)
+ P(X = 6)
31
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3 Discrete random variables
( ) ( 18 ) + ( 78 ) ( 18 ) + ( 78 ) ( 18 )
7 1
+ (8) (8)
7
= 8
2
3
b P(X = n) = (1 − s)n − 1s
4
c P(X = 2) = 0.21
5
(1 − s)s = 0.21
s − s2 = 0.21
= 0.317 (3 d.p.)
s2 − s + 0.21 = 0
d i P(X < 4) = P(X = 1) + P(X = 2) + P(X = 3)
s = 0.7 or 0.3
= 0.4 + 0.6(0.4) + 0.62(0.4)
= 0.784
ii P(X = 5) = 0.64 × 0.4 = 0.051 84
iii P(X > 5) = (1 − 0.4)5 = 0.077 76
e
i P(X  3) = P(X = 1) + P(X = 2) + P(X =3)
Since the probability of a person answering
the phone on the first call is greater than 0.5,
s = 0.7.
9
= 0.16 + (1 − 0.16)(0.16)
+ (1 − 0.16)2(0.16) = 0.407 (3 d.p.)
ii P(X > 6) = (1 − 0.16)6 = 0.351 (3 d.p.)
iii P(X  4) = P(X > 3) = (1 − 0.16)3 = 0.593 (3 d.p.)
2
X ~ Geo
a X ~ Geo  1 
 12 
P ( X 20 ) = P(X > 19) =
x
 11  = 0.1
 12 
(101 )
x ln
P(X = 6) =  1 − 1   1  = 0.0590 (3 s.f.)
 10   10 
4
X ~ Geo(0.5)
P(X = 4) = 0.53 × 0.5 = 0.0625
a X ~ Geo(0.25)
P(X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
She must collect at least 27 beetles.
10 a Y ~ Geo  1 
 20 
( 1920 ) ( 201 ) = 0.031 5
b P (Y > 25) = ( 19 ) = 0.277
20
c P (Y 15) = 1 − ( 19 ) = 0.537
20
P (Y = 10 ) =
15
b P(X  7) = P(X > 6) = (1 − 0.25)6 = 0.178
P(X = 6) = 0.655 × 0.35
Exercise 3.4B
= 0.0406 (3 s.f.)
b P(X  10) = P(X > 9)
= 0.659 = 0.0207 (3 s.f.)
6
X ~ Geo(0.4)
There is time for three attempts in 1 h 30 min
(90 ÷ 25 = 3.6)
P(X  3) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.4 + 0.6 × 0.4 + 0.62 × 0.4 = 0.784
7
8
1
a
 10
i P(X = 2) =   × 0.22 × 0.88 = 0.302
 2
 10
ii P(X = 5) =   × 0.25 × 0.85 = 0.0264
 5
 10
iii P(X = 10) =   × 0.210 × 0.80 = 1.02 × 10−7
 10
b iP(X = 2) =
 7
× 0.82 × 0.25 = 0.004 30
 2
3
ii P(X = 5) =
 7
× 0.85 × 0.22 = 0.275
 5
5
iii Since the binomial distribution is discrete,
()
4 1
P(X = 4) = ( ) ( ) = 0.1024
5 5
4
b P(X > 5) = ( ) = 0.328 (3 s.f.)
5
1
a X ~ Geo 5
a Each phone call is independent.
The probability of the phone calls getting
answered is constant.
32
9
25
= 0.684
a X ~ Geo(0.35)
11
= ln0.1
12
x = 26.46
=0
.25 + 0.75 × 0.25 + 0.752 × 0.25
+ 0.753 × 0.25
5
19
b P(X > x) = 0.1
5
3
(1211 ) = 0.191
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
=
 7
 7
× 0.80 × 0.27 +   × 0.81 × 0.26
 0
 1
 7
+   × 0.82 × 0.25 = 0.004 67
 2
©HarperCollinsPublishers 2018 Cambridge International AS & A Level Mathematics: Probability & Statistics 1 9780008257767
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WORKED SOLUTIONS
 20
i P(X = 18) =   × 0.12518 × 0.8752
 18 
c
e
 17
i P(X  2) =   × 0.160 × 0.8417
 0
= 8.08 × 10−15
ii From cumulative binomial probabilities,
 17
+   × 0.161 × 0.8416
 1
P(X  3) = P
(X = 0) + P(X = 1) + P(X = 2)
+ P(X = 3)
 17
+   × 0.162 × 0.8415
 2
() ()
0
20
 20
1
7
= ×
×
8
8
0
×
( 78 )
19
3
= 0.473
 17
ii P(X = 5, 6, 7) =   × 0.165 × 0.8412
19
2
18
1
2
18
1
19
2
18
1
 5
20
20
20
20
20
20
 
 
1
7
1 × 77 +   × 11 × 77 +   × 11
+  ×
×
+  ×
× 8 +   × 8 × 8 +   × 8 17
8
8
8
 17
8
8
8
8


 1
 2
 22 
 11  +
× 0.166 × 0.8411 +   × 0.167 × 0.8410


 7
3
17
 6
 20
1
7
+  ×
×
= 0.120
8
8
 3
() ()
() ()
() ()
() ()
()
iii P(2 < X  15) = P(X  15) − P(X  2)
= 0.765
  17
iii P(3 < X  18) =
P(X  18) − P(X  3)
16
1  17 
17
0
= 1 −   16 × 0.16 × 0.84 +  17 × 0.16 × 0.84 
= (1 − P(X = 19 or 20)) − 0.765 33
19
1
20
  20
 20
1
7
1 

 17 
1 −    × 8 × 8 +   × 8  17 
 20
× 0.1617 × 0.840  − 0.473…
× 0.1616 × 0.841 + 
  19 



 17 
  16 

− 0.765 33
= 0.527
= 0.235
2 Number of trials is eight, therefore n = 8.
d iSince the binomial distribution is discrete,
1
P(X < 4) = 1 − P(X = 4 or 5)
Probability of rolling a six is 1 , therefore p = .
6
6
  5

 5
Since
there
are
a
set
number
of
independent
5
0
4
1
=1−
  × 0.4 × 0.6 +  4 × 0.4 × 0.6 
trials with a consistent probability, the problem
  5

can be modelled using a binomial distribution.
= 0.913
() ()
()
Let X be the number of sixes rolled.
Alternatively, P(X = 0, 1, 2 or 3)
 1
Therefore, X ~ B 8, 
 6
5
 5
× 0.41 × 0.64
=   × 0.40 × 0.65 +
 1
 0
+
5
 5
× 0.42 × 0.63 +   × 0.43 × 0.62
 2
 3
= 0.913
 5
ii P(X = 5) =   × 0.45 × 0.60 = 0.010 2
 5
iii P(0  X < 3)
= P(X < 3)
= P(X  2) = P(X = 0) + P(X = 1) + P(X = 2)
 5
=   × 0.40 × 0.65
 0
+
 5
× 0.41 × 0.64
 1
 5
+   × 0.42 × 0.63
 2
= 0.683
() ()
3
5
 8
P(X = 3) =   × 1 × 5 = 0.104
6
6
 3
 10
× (s)6 × (1 − s)4
 6 
3
P(X = 6) =
4
a P(X < 3) = P(X  2)
 12
 12
0
12
1
11  12
=   × 0.25 × 0.75 +   × 0.25 × 0.75 +   × 0.
 0
 1
 2
 12
+   × 0.251 × 0.7511
 1
 12
+   × 0.252 × 0.7510
 2
= 0.391
 12
b P(X  9) =   × 0.259 × 0.753
 9
 12
+   × 0.2510 × 0.752
 10
33
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3 Discrete random variables
+
Therefore, X ~ B(10, 0.25)
 12
× 0.2511 × 0.751
 11
 10
P(X = 3) =   × (0.25)3 × (0.75)7
 3
 12
+   × 0.2512
 12
= 0.250 (3 s.f.)
b P(X  3) = 1 − P(X  2)
= 1 − 0.5256
= 0.474 (3 s.f.)
= 0.000 392 (3 s.f.)
5
a
Number of students selected is 30, therefore
n = 30.
Probability of a student asking for paper is 3
out of 5, therefore p = 0.6.
7
a Number of students is 20, therefore n = 20.
Probability of having a birthday in January is
1 , therefore p = 1 .
12
12
Since there is a set number of independent
students and a fixed probability of having
a birthday in January, the problem can be
modelled as a binomial distribution.
Since there is a set number of independent
trials and a fixed probability of success, the
problem can be modelled as a binomial
distribution.
Let X be the number of students who ask for
paper.
Let X be the number of students with
birthdays in January.
1
Therefore, X ~ B 20, 

12 
Therefore, X ~ B(30, 0.6)
 30
P(X = 10) =   × (0.6)10 × (0.4)20
 10 
( ) ( )
1
19
20
P(X = 1) =   × 1 × 11 = 0.319 (3 s.f.)
 1  12
12
= 0.002 00 (3 s.f.)
b P(X < 25) = 1 − P(X  25)
b P(X  4) = 0.978 (3 s.f.)
  30

 30 
 30 
 30 
 30
 30
25
5
26
4
= 1 –    × 0.6 × 0.4 +   × 0.6 × 0.48 + a
× 0.6 27of
× 0customers
.43 +   chosen
× 0.6 28 ×is05,
.4
4 2therefore
+   × 0.6 29 × 0.4 +   × 0.630 
Number


 28
 26
 27
 29
 30
  25

n = 5. Probability of being in the 18–25 age
group is 0.4.
Since
there is a set
  30
of
 30

 30number

25
5  30 
26
4  30 
27
3  30 
28
29
30
4 2 +  customers
0
6
0
4
0
6
×
.
×
.
+
.
× 0.4
×
  25 × 0.6 × 0.4 +  26 × 0.6 × 0.4 +  27 × 0.6 × 0.4 +  28 × 0.6independent

and
a
fixed
probability,
 30
 
 29
 
 
 

the problem can be modelled as a binomial
distribution

 30 
 30
 30 
 30B(5, 0.4).
 30 
0.6 25 × 0.45 +   × 0.6 26 × 0.4 4 +   × 0.6 27 × 0.43 +   × 0.6 28 × 0.4
4 2 +   × 0.6 29 × 0.4 +   × 0.630 
Let X be the
30number of customers in the age
 28
 29
 27
 26
group 18–25. Therefore:

 30
26
4  30 
27
3  30 
28
2  30 
29
30P(X < 2) = P(X = 0) + P(X = 1)
0.6 × 0.4 +   × 0.6 × 0.4 +   × 0.6 × 0.4
4 +   × 0.6 × 0.4 +   × 0.6 
 29
 28
 30
 27

 5
0
5  5
1
4
=   ( 0.4 ) ( 0.6 ) +   ( 0.4 ) ( 0.6 )

30
30



30



 1
 0
4 2 +   × 0.6 29 × 0.4 +   × 0.630 
0.6 27 × 0.43 +   × 0.6 28 × 0.4
 30
 29
 28

= 0.337

 30
b The binomial distribution will not be
+   × 0.630 
 30

suitable. Customers chosen from the same
6
= 1 − 0.005 659 = 0.994 (3 s.f.)
a
Number of flights in the winter is 10, therefore
n = 10.
Probability of poor visibility is 25%,
therefore p = 0.25.
Since there are a set number of independent
trials with a fixed probability of success,
the problem can be modelled as a binomial
distribution.
Let X be the number of poor visibility
encounters.
9
table are not independent of each other;
they are more likely to be in the same age
group.
A packet contains 10 gingerbread men biscuits,
therefore n = 10. Probability of being broken is
0.03. Since there are a set of independent trials
with a fixed probability, this can be modelled as a
binomial distribution, X~B (10, 0.03).
10
P(X = 2) =   ( 0.03)2 ( 0.97 )8 = 0.031 7
 2 
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WORKED SOLUTIONS
10 A box contains 10 duck eggs, therefore n = 10.
Probability of any egg being broken is 0.1. Since
there is a set number of independent duck eggs
and a fixed probability, this problem can be
modelled as a binomial distribution, X~B (10, 0.1).
ii Var(X) = np(1 − p)
= 5 × 0.1 × (1 − 0.1)
3
P(X  2) = 1 – P(X  1)
= 0.264
a E(X) =
4
=3
Var(X) = np(1 − p)
= 30 × 0.1 × (1 − 0.1)
= 2.7
1
1
b E(X ) = =
p 0.45
b P(μ  X < μ + σ) = P(3  X < 3 + 1.643)
= P(X = 3) + P(X = 4)
20
9
=
= 2.22 (3 s.f.)
1
1
c E(X) = =
p
3
7
7
=
3
()
= 2.33 (3 s.f.)
2
a E(X) = np
= 30 × 0.1
1= 1
p 0.3
= 10
3
= 3.33 (3 s.f.)
=
()
1

b Y ~ B 42, 

6
E(Y) = np
1
= 42 ×
6
=7
  10
  10
0
10 
1
9 
   ( 0.1) ( 0.9)  +    ( 0.1) ( 0.9)  
  1 

  1  = 1 – 0.736 1
1
= 0.45
1
a X ~ Geo
6
E(X ) = 1 = 1 = 6
p  1
 6
  10
0
10    10 
1
9 
= 1 – 
 0  ( 0.1) ( 0.9)  +   1  ( 0.1) ( 0.9)  
 
 

Exercise 3.5A
a i E(X) = np
= 10 × 0.2 = 2
ii Var(X) = np(1 − p)
= 10 × 0.2 × (1 − 0.2)
= 1.6
b i E(X) = np
= 150 × 0.3 = 45
ii Var(X) = np(1 − p)
= 150 × 0.3 × (1 − 0.3)
= 31.5
c i E(X) = np
= 20 ×
3
3
= 7.5
8
ii Var(X) = np(1 − p)
3
= 20 × × (1 − 3 )
8
8
= 4.687 5
d i E(X) = np
= 5 × 0.1 = 0.5
 30
× 0.13 × 0.927
 3 
+
 30
× 0.14 × 0.926
 4 
= 0.413
5
E(X) = np
= 25 × 0.6
= 15
Var(X) = np(1 − p)
= 25 × 0.6 × (1 − 0.6)
=6
So σ = 6 = 2.449 4… = 2.45 (3 s.f.)
P(μ − σ < X < μ + σ) = P(15 − 2.449 5 < X < 15 + 2.449 5)
= P(12.550 5 < X <17.449 5)
= P(X = 13) + P(X = 14) + P(X = 15)
+ P(X = 16) + P(X = 17)
 25
=   × 0.613 × 0.412
 13 
 25
+   × 0.614 × 0.411
 14 
 25
+   × 0.615 × 0.410
 15 
 25
+   × 0.616 × 0.49
 16 
 25
+   × 0.617 × 0.48
 17 
= 0.693 (3 s.f.)
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3 Discrete random variables
6
E(X) = np = 2.5
Var(X) = np(1 − p) = 1.875
b To obtain the second $1 coin, there are only 14
coins in the bag. Therefore the probability of
getting a $1 coin is 4 . Let Y be the number of
14
extra selections; this again can be modelled as
geometrical distribution Y~Geo 4 .
14
1 = 7 or 3.5 selections
E(X) =
4 2
14
(1)
(2)
( 2) : 1 − p = 1.875
2.5
(1)
( )
1 − p = 0.75
p = 0.25, and by substitution, n = 10
2.5
Or E(X) = np = 2.5 so n = p
(1)
Var(X) = np(1 − p) = 1.875
(2)
Substitute (1) into (2):  2.5  p(1 − p) = 1.875
 p 
10 The probability of getting a chocolate egg with a
1
limited edition toy is . Let X be the number of
12
chocolate eggs up to and including the first limited
edition toy, this problem can be modelled as
1
.
geometrical distribution, X ∼ Geo
12
E(X) = 1 = 12 eggs
1
12
Expected cost: $1.20 × 12 = $14.40
2.5(1 − p) =1.875
1.875
1−p=
2.5
7
( )
= 0.75
p = 0.25, n = 10
a X ~ B n, 13
20
(
)
E(X) = 39
13
n×
= 39
20
20
n = 39 ×
13
= 60
39
21
 60
b P(X = 39) =   ×  13  ×  7 
 20 
 39  20 
= 0.107 (3 s.f.)
8
Exam-style questions
1
P(X = 1) = 1 – (0.3 + 0.4 + 0.05)
= 0.25
b E(X) = 0 × 0.3 + 1 × 0.25 + 2 × 0.4 + 3 × 0.05
= 1.2
2
a X ~ B(8, 0.1)
 20
a P(T = 3) =   × 0.43 × 0.617
 3
= 0.012 3
E(X) = 8 × 0.1 = 0.8, so about 1 plate is expected to
be damaged in each box.
b P(T  3) = P(T = 0) + P(T = 1) + P(T = 2) + P(T = 3)
b P(X  1) = 1 − P(X = 0)
 20
=   × 0.40 × 0.620
 0
  8

= 1 −    × 0.10 × 0.98 
  0

 20
+   × 0.41 × 0.619
1
= 0.570 (3 s.f.)
20
+   × 0.41 × 0.618
 2 
c Y ~ B(40, 0.569 5…)
E(Y) = 40 × 0.569 5…
= 22.8 (3 s.f.), so 23 boxes are expected to
contain one or more damaged plates.
+  20 × 0.43 × 0.617
 3 
d There can only be an integer number of boxes.
9
a 0.3 + P(X = 1) + 0.4 + 0.05 = 1
= 0.016 0
aThere are 15 coins in total, five of them
c E(T) = 20 × 0.4
=8
are $1 coins. Therefore the probability of
5
selecting one $1 coin is . Let X be the
15
number of selections up to and including
the first $1 coin, this can be modelled as
geometrical distribution X~Geo 5 .
15
1 = 3 selections
E(X) =
5
15
( )
Var(T) = 20 × 0.4 × 0.6
= 4.8
3
a The trials need to be independent of each other.
The probability of a success needs to be constant.
There must be only two possible outcomes
with fixed probabilities.
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WORKED SOLUTIONS
b P(X = 3) = (1 − 0.4)2 × 0.4
 15
b P(X = 15) =   (0.1)15(0.9)0 = 1 × 10−15
 15
= 0.144
c P(X > 3) = (1 − 0.4)3
= 0.216
4
c E(X) = 15 × 0.1 = 1.5
a If S = 4 then you can have 1 and 3 or 3 and 1.
7
a P(3 fails, then 1 success) = 0.73 × 0.3 = 0.103
b P(1st success after 4th throw) = 0.74 = 0.240
so P(S = 4) = 1 × 2 + 2 × 1
4 3 4 3
1
=
3
1 1
b P(S = 1) = 2 ×
4 3
c P(2nd success on 4th throw) = P(1 success in
1st 3 throws, then 1 success on 4th throw)
  3

=   (0.3)1(0.7)2  × 0.3
1




( )
2 1
P(S = 3) = 2 ( × )
4 3
= 0.132
d P(Lucas first scores on his 2nd shot)
= 0.7 × 0.8 × 0.3 = 0.168
P(S = 6) = 2 1 × 1 
 4 3
s
1
3
4
6
P(S = s)
1
6
1
3
1
3
1
6
e P(Stefan first scores on his 7th attempt)
= (0.7 × 0.8)6 × 0.7 × 0.2 = 0.00432
8
a E(R) = −2a + 3(1−a)
= 3 – 5a
Var(R) = (−2)2a + 32(1 – a) – E2(R)
1
1
1
1
c E(S) = 1 × 6 + 3 × 3 + 4 × 3 + 6 × 6
= 4a + 9(1 – a) – (3 – 5a)2
= 3.5
= 25a – 25a2
1
1
1
1
Var(S) = 12 × 6 + 32 × 3 + 42 × 3 + 36 × 6 – 3.52
= 2.25
5
3
= 25a(1 – a)
b E(R) = 2
1
5
X ~ B(10, 0.21)
Therefore: 3 – 5a = 2 ⇒
a=
 10
a P(X = 2) =   (0.21)2 (0.79)8 = 0.301 (3 s.f.)
 2
Var(R) = 25 × 4 = 4 ⇒
5 5
σ = 4=2
b P(X  4) = 1 – P(X  3)
9
Y ~ B(30, 0.05)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]
 30
a P(X = 2) =   (0.05)2(0.95)28 = 0.259
 10

 2
10
10
10






= 1 −   (0.21)0(0.79)10 +   (0.21)1(0.79)9 +   (0.21)2(0.79)8 +   (0.21)3(0.79)7 
3 3) = 1 − P(X  3)
 2
>
 1
b P(X
 0 

 10
 10
 10
0.21)0(0.79)10 +   (0.21)1(0.79)9 +   (0.21)2(0.79)8 +   (0.21)3(0.79)7 
 2
 1
 3

= 1 − 0.860 8
= 0.139 (3 s.f.)
= 1 − [P(X = 0) + P(X = 1) + P(X = 2)
+ P(X = 3)]
 30
 30
 30
0
30
1
29
= 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   (
0
1




 2

c P(4 < X < 9) = P
(X = 5)+ P(X = 6) + P(X = 7)
+ P(X = 8)
 30
0
30  30 
1
29  30 
2
28 
= 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) 
1
2
0






  10

 10

 10
 10
=   (0.21)5(0.79)5 +   (0.21)6(0.79)4 +   (0.21)7(0.79)3 +   (0.21)8(0.79)2 
 7 
 8 
 30 6  0

30  30
1
29  30
2
28  30 
 5 
= 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   (0.05)3 (0.95)27 
2
0
1
3
 
 
 
 10
 
 10
 10
  10

=   (0.21)5(0.79)5 +   (0.21)6(0.79)4 +   (0.21)7(0.79)3 +   (0.21)8(0.79)2 
5
6
7
8









 30

 30
0
30  30 
1
29  30 
2
28
= 1 −   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   ( 0.05) ( 0.95) +   (0.05)3 (0.95)27 
1
 2
 3

 0=0.039 9
6 X ~ B(15, 0.1)
= 1 − 0.939 2 …
 15
= 0.060 8
a P(X = 3) =   (0.1)3(0.9)12 = 0.129
 3
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3 Discrete random variables
c Y ~ B(20, 0.06077)
Blue beads:
 20
P(Y = 5) =   (0.06077)5(0.93923)15
 5
= 0.005 02 (3 s. f.)
( )( )
6
4
 10
9 = 0.074 4 …
P(X = 6) =   5
 6  14 14
d E(X) = 20 × 0.06077 = 1.22


10 X ∼ B5, 1 
 7
X ~ B 10, 5 
 14 
P(exactly six beads of same colour)
= 0.000 96… + 0.007 7… + 0.029 7… + 0.074 4…
P(exactly six beads of same colour) = 0.113(3 s.f.)
14 Let X be the number of bullets that hit the target.
X ~ B (6, 0.7)
( )( )
3
2
 5
P(3 children) =   1 6 = 0.021 4 …
 3 7 7
P(X  3) = 1–P(X = 0) – P(X = 1) – P(X = 2)
2
3
 5
P(2 children) =    1   6  = 0.128 5 …
 2  7   7 
P(3 or 2 children develop the condition)
= 0.021 4 … + 0.128 5 … = 0.150


11 X ∼ B6, 4 
 9
 6
0
10  6 
1
9  6
2
8
= 1 −   ( 0.7 ) ( 0.3) +   ( 0.7 ) ( 0.3) −   ( 0.7 ) ( 0.3) 
 2
 1
 0

 6
0
10  6 
1
9  6
2
8
+   ( 0.7 ) ( 0.3) +   ( 0.7 ) ( 0.3) 
= 1 −   ( 0.7 ) ( 0.3) −
 2
 1
 0

= 1 – 0.070 47
( )( )
= 0.930
15 aThe spins are independent of each other and
the order of the outcomes does not matter.
4
2
 6
a P(X = 4) =   4 5 = 0.181
 4 9 9
b P(X  4) = P(X = 4) + P(X = 5) + P(X = 6)
( )( )
( )( )
()
4
2
5
1
6
 6
 6
 6
=   4 5 +  4 5 +  4
9
9
9
9
9
 5
 6
 4
= 0.181 + 0.058 + 0.008
= 0.246
1
1
b P( R ) = 1 , P( B ) = , P(Y ) =
6
3
2
x
2R
1× 1 = 1
2 2 4
1R and 1B
1× 1×2 1
= 3
2 3
1R and 1Y
1× 1×2 1
=6
2 6
2B
1× 1 1
3 3=9
1B and 1Y
1× 1×2 1
=9
6 3
2Y
1× 1
1
6 6 = 36
12 a X ~ B(10, 0.01)
P(X  1) = 1 − P(X = 0)
=1−
 10
(0.01)0(0.99)10
 0 
= 0.095 6
b Y ~ B(5, 0.09562)
 5
P(Y = 3) =
(0.09562)3(0.90438)2 = 0.00715
 3
( )
 10 1 6
P(R = 6) =
= 0.000 96 …
 6  ( 7 ) ( 7 )
3
Yellow beads: Y ~ B(10, )
14
 10
P(Y = 6) =   ( 3 ) ( 11 ) = 0.007 7 …
 6  14 14
2
Green beads: G ~ B(10, 7 )
13 Red beads:
1
R ~ B 10,
7
6
4
6
( )( )
4
6
4
 10
P(G = 6) =   2 5 = 0.029 7 …
7
7
 6
P(X = x)
c Let YB be the random spin that landed on
yellow from the biased spinner and YF be the
random spin that landed on yellow from the
fair spinner.
P(Y B ) = 1 and P(Y F ) = 1
3
6
Therefore, the probability of landing on
yellow is: 1 × 1 + 1 × 1 = 1
2 6 2 3 4
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WORKED SOLUTIONS
()
1
19 a The geometric distribution Geo 5 .
 1
d Y ~ Geo 
4
( ) ( 15 ) = 0.081 9
b P(X = 5) = 4
5
Bob has at least three attempts before he
wins, that is:
( ) = 169
P(Y  3) = 3
4
3
4
1
4
2
 4
c P( X 5) =   = 0.410
 5
16 The number of weeks from July to October is
16, therefore n = 16. The probability of rainfall
is 0.35. Let X be the number of weeks that have
rainfall. This problem can be modelled as binomial
distribution X ~ B(16, 0.35).
d Assume that the outcome of each kick is
independent of the results of any other kicks.
Assume that the probability of scoring a goal
remains constant.
20 a X ~ Geo(0.04)
P(X  3) = 1 − P(X > 3) = 1 − 0.963 = 0.115
 16
 16
0
16  16 
1
15  16 
2
3
13 
 )114 +
= 1 −   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +  b( 0.E(X)
35) (=
0.65
0.35) ( 0.65) 
(
=
25




 1
 2
 0.04   3 
 0 

P(X < 25) = 1 − P(X > 24)
 16

0
16  16 
1
15  16 
2
14  16 
= 1 −   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0.35 )3(0.65)13 = 1 − (0.96)24
1
2
3
0









 0.625
=
6
c Y ~ B(10, 0.115)
0
16  16 
1
15  16 
2
14  16 
3
13 
 ( 0.35) ( 0.65) +  1  ( 0.35) ( 0.65) +  2  ( 0.35) ( 0.65) +  3  ( 0.35) ( 0.65) 
E(Y) = 10 × 0.115 = 1.15

P(X > 3) = 1 – P(X  3)
Var(Y) = 10 × 0.115 × (1 − 0.115) = 1.02
 16
3
13  16 
34
13
16   16 
4
16 
+   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) +   ( 0.35) ( 0.65) 
 3
 34 
 4

Assume that whether the machine is stopped on
each run is independent of the outcomes of the
other runs.
= 1 – 0.133 86 = 0.866 (3 s.f.)
Assume that the probability of rainfall each week is
constant.
Assume that each time it rains is independent of
whether it rains in the other weeks.
17 aThere are nine single positive digits, which are
1, 2, 3, 4, 5, 6, 7, 8, 9
Therefore, probability of getting a six is P(6) = 1 .
9
or
Assume that the probability that the
machine is stopped remains constant.
Mathematics in life and work
X ~ B(100, 0.999)
1
 100
P(X = 99) = 
(0.999)99(0.001)1 = 0.0906
 99 
2
P(X  98) = P(X = 98) + P(X = 99) + P(X = 100)
b Let X be the number of digits that the
computer program generated up to and
 1
including the first 6. X ~ Geo 
 9
12
8


P(X 13) =   = 0.243
 9
1
c P(N=1) = 0.2433 × 9 = 0.0270
 100
=
(0.999)98(0.001)2
 98 
 100
+
(0.999)99(0.001)1
 99 
 100
+
(0.999)100(0.001)0
 100
8
 8 1
d P(N = 9) = 0.2433 ×   × = 0.010 5
 9 9
= 0.999 8
18 a X ~ Geo (0.65)
b P(X = 1) = 0.65
c
P(X  5) = 1 – P(X > 5) = 1 – 0.355 = 0.995
d 200 × 0.65 = 130 people
3
P(more than 3 inaccurate bags)
= P(X  96) = 1 − P(X  97)
= 1 − [P(X  98) + P (X = 97)]
39
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3 Discrete random variables
 100
(0.999)97(0.001)3
= 1 − 0.999 849 62 − 
 97 
= 1 − 0.999 849 62 − 0.000 146 75
= 3.63 × 10−6
4
P(more than 4 inaccurate bags)
= P(X  95) = 1 − P(X  96)
= 1 − [P(X  98) + P (X = 97) + P (X = 96)]
= 1 − 0.999 849 623 6 − 0.000 146 744 7
 100
−
(0.999)96(0.001)4
 96 
= 1 − 0.999 849 623 6 − 0.000 146 744 7
− 0.000 003 562 1
= 6.96 × 10−8
40
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4
WORKED SOLUTIONS
4 Normal distribution
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering
the question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Where values from the Cambridge Assessment International Education statistical tables are used, the same level of
accuracy has been used in workings unless stated otherwise.
Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication
of the answer.
Prerequisite knowledge
1
6
a Mean = 3.75
b Standard deviation = 2.12
b The mean of random variable B is 4.8.
c New mean = 3.32, new standard deviation = 2.24
2
a E(Y ) = 11
3
σ = 1.31 (3 s.f.)
b P(Y > m + σ ) = P(Y > 4.98) = 1
6
Exercise 4.1A
1
2.4
7
a The distribution is symmetrical about the mean.
c The total area below the curve is 1.
d The distribution is defined by two parameters:
the mean and the standard deviation.
3
4
Standard deviation is equal. Mean is different for
the two distributions.
17
a 50%
b
50
c 47.5%
d 47.5% + (99.75 ÷ 2) = 779
800
Curve A has a higher mean compared to curve B.
3.6
4.8
6
7.2
8.4
9.6
10.8
a Always true.
bNever true. The point of inflexion is located one
standard deviation on either side of the mean.
cNever true. (It is possible but the probability is 0.)
b The mode, median and mean are all the same.
2
a Random variable A: The red curve
Random variable B: The blue curve
Random variable C: The black curve
d Always true.
8
a Mean = 2.5 mm
Standard deviation = 0.1 mm
b 68% of the screws have a diameter measurement
within one standard deviation.
9
a68% of the data is located within one standard
deviation.
167 ± 1.8 = (165.2, 168.8)
Curve A has a smaller standard deviation
compared to curve B.
5
X
16%
16%
165.2 cm
Y
42.5
Standard deviation of blue curve, X, is 1.5
Standard deviation of red curve, Y, is 2.5
167.0 cm
168.8 cm
bTo find how many standard deviations away
from the mean, 169.7 − 168 = 1.5 standard
1.8
deviations, this is about 6.75%  16% − 2.5%  on


2
the right of 1σ.
Probability = 0.025 + 0.0675 = 0.0925, so about
9.25%.
41
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4 NORMAL DISTRIBUTION
6
Adam is wrong because both random variable X
and random variable Y have the same variance,
1.52 = 2.25
16%
Billy is wrong because it is possible to have two
different random variables which have the same
mean and variance.
2.5%
m
s
Neither of them were right.
2s
7
10
i Graph b and graph c are normal distribution
curves because they show:
› symmetry about the centre
› a bell-shaped curve.
ii Graph b – N(21, 1.52)
16%
Graph c – N(16.7, 2.32)
8
m
16% of the students failed the exam, which means
that these students did not pass the bench mark.
From the graph, it shows that this is 1 standard
deviation to the left of the mean.
Therefore, the pass mark for the test = 68 – 15 = 53
b Sometimes true. Many continuous random
variables can be modelled by a normal
distribution, e.g. heights of people, blood
pressure, errors in measurements, etc.
However, not every continuous random
variable is normally distributed.
Exercise 4.2A
1
2
3
4
A ~ N(3, 4), B ~ N(2, 3²)
Let A represent Year 1 scores and B represent Year 2
scores.
A ~ N(0, 5), B ~ N(−1, 2²)
The ‘N’ and ‘X’ are in incorrect places; the mean
should be 7 and the standard deviation should
be 11.
X ~ N(7, 11²)
a Student A: mean = 5, variance = 49, standard
deviation = 7
a Sometimes true. In the case where t = 1, the
variance 12 is equal to the standard deviation 1
and so the statement is true. Otherwise, the
statement is untrue.
9
a Experiment B
b Experiment B has the shortest average
reaction time; it also has the smallest
variance. This indicates that the reaction
times are less varied.
10 a Let X be the mass of mixed nuts, X ~ N(50, 2.32)
b
Student B: mean = 6, variance = 81, standard
deviation = 9
16%
16%
b
A
B
47.7 g
50 g
a
0
52.3 g
Exercise 4.2B
1
a
5
X
Z
Y
m
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4
WORKED SOLUTIONS
b
5
a
Z
0
0
c
1.064
P(Z > 1.064) = 1 − Φ(1.064)
= 1 − 0.8563
= 0.1437
a
0
b
Z
6
d
–2.872
a
b 0
0
P(Z < −2.872) = Φ(−2.872)
Z
= 1 − Φ(2.872)
e
= 1 − 0.9979
= 0.0021
7
0
2
a
Z
‘N’ and ‘Z’ are written in the incorrect positions.
The mean should equal 0. The standard deviation
should equal 1.
Z ~ N(0, 1²)
–1.326
0
P(Z > −1.326) = 1 − Φ(−1.326)
= 1 − [1 − Φ(1.326)]
3
= Φ(1.326)
= 0.9076
8
0
1.423
P(Z < 1.423) = Φ(1.423)
= 0.9226
4
0
1.1
2.1
Z
P(1.1 < Z < 2.1) = Φ(2.1) − Φ(1.1)
= 0.9821 − 0.8643
= 0.1178
0
0.87
Z
P(Z < 0.87) = Φ(0.87)
= 0.8078
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4 NORMAL DISTRIBUTION
P(0.24 < Z < 1.102) = Φ(1.102) – Φ(0.24)
= 0.8647 – 0.5948 = 0.2699
9
Exercise 4.3A
1
a P(0 < Z < 1.2) = Φ(1.2) − Φ(0)
= 0.8849 − 0.5
–1.325
0
= 0.3849
1.218
b P(−2.25 < Z < 0) = Φ(0) − Φ(−2.25)
P(−1.325 < Z < 1.218) = Φ(1.218) − Φ(−1.325)
= Φ(0) − [1 − Φ(2.25)]
= Φ(1.218) − [1 − Φ(1.325)]
= Φ(1.218) + Φ(1.325) − 1
= Φ(0) + Φ(2.25) − 1
= 0.8884 + 0.9074 − 1
= 0.5 + 0.9878 − 1
= 0.7958
= 0.4878
c P(−0.5 < Z < 1.5) = Φ(1.5) − Φ(−0.5)
10
= Φ(1.5) − [1 − Φ(0.5)]
= Φ(1.5) + Φ(0.5) − 1
= 0.9332 + 0.6915 − 1
= 0.6247
d P(Z > 2) = 1 − Φ(2)
0
–2.651 –1.43
= 1 − 0.9772
= 0.0228
P(−2.651 < Z < −1.43)
(
= [1− Φ(1.43)] − [1 − Φ(2.651)]
35 − 30
10
= P(Z < 0.5)
= Φ(2.651) − Φ(1.43)
= Φ(0.5)
= 0.9960 − 0.9236
= 0.6915
= Φ(−1.43) − Φ(−2.651)
= 0.0724
2
a P(X < 35) = P Z <
)
b P(X > 38.6) = 1 − P(X < 38.6)
(
11
=1−P Z <
38.6 − 30
10
)
= 1 − P(Z < 0.86)
= 1 − Φ(0.86)
= 1 − 0.8051
= 0.1949
–0.541
P( Z < – 0.541) = P( Z > 0.541)
= 1 − P(Z < 0.541)
= 1 − 0.7057
= 0.2943
c P(X > 20) = 1 − P(X < 20)
20 − 30
=1−P Z <
10
(
)
= 1 − P(Z < −1)
= 1 − Φ(−1)
12
= 1 − [1 − Φ(1)]
= Φ(1)
= 0.8413
0
1.102
d P(35 < X < 40) = P(X < 40) − P(X < 35)
40 − 30
35 − 30
=P Z <
−P Z <
10
10
(
) (
)
44
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WORKED SOLUTIONS
4
= P( Z > 2.938 )
= P(Z < 1) − P(Z < 0.5)
= Φ(1) − Φ(0.5)
= 1 – Φ (2.938)
= 1 – 0.9883
= 0.0017
= 0.8413 − 0.6915
= 0.1498
10 − 10.8  = P Z −0.5 = 1 − 0.5

e P(15 < X < 32) = P(X < 32) − P(X < 15)
b P( X 10 ) = P  Z (
)
( 5) = 1 − 0.691

1.6 
32 − 30
15
−
30
=P Z <
− P Z <  10 − 10.8  = P Z −0.5 = 1 − 0.5
10P( X 10 ) = P  Z10
(
)
( 5) = 1 − 0.6915 = 0.3085
1.6 
= P(Z < 0.2) − P(Z < −1.5)
c P(9.2 < X < 11.2) = P( X < 11.2) – P( X < 9.2)
) (
(
)
(
= Φ(0.2) − Φ(−1.5)
=P Z <
= Φ(0.2) − [1− Φ(1.5)]
= Φ(0.2) + Φ(1.5) − 1
= P(Z < 0.25) – P( Z < – 1)
= 0.5793 + 0.9332 − 1
= Φ (0.25) – (1 – Φ (1))
= 0.5125
= 0.5987 – (1 – 0.8413)
f P(17 < X < 19) = P(X < 19) − P(X < 17)
= P  Z < 19 − 30  − P  Z < 17 − 30 


10 
10 
a X ~ N(177, 7²)
(
= P(Z < −1.1) − P(Z < −1.3)
= Φ(−1.1) − Φ(−1.3)
= Φ(−1.5)
= [1 − Φ(1.1)] − [1 − Φ(1.3)]
= 1 − Φ(1.5)
= Φ(1.3) − Φ(1.1)
= 1 − 0.9332
= 0.9032 − 0.8643
= 0.0668
P(X < 4.5) = P Z <
(
b P(X > 8) = 1 − P(X < 8)
(
=1−P Z < 8−6
1
)
= 1 − 0.9772
= 0.0228
= 1 − 0.7623
c P(4.5 < X < 8) = P(X < 8) − P(X < 4.5)
= 0.2377
(
180 − 177
7
= 1 − P(Z < 0.429)
=1−P Z <
)
= 1 − Φ(2)
= 1 − Φ(0.714)
b P(X > 180) = 1 − P(X < 180)
)
= 1 − P(Z < 2)
= Φ(−0.714)
= 0.9772 − 0.0668
)
= 1 − Φ(0.429)
= 0.9104
6
a X ~ N(60, 4²)
(
55 − 60
4
= P(Z < −1.25)
P(X < 55) = P Z <
= 1 − 0.6660
= Φ(−1.25)
= 0.3340
= 1 − Φ(1.25)
c P(172 < X < 180) = P(X < 180) − P(X < 172)
= 0.6660 − 0.2377
= 0.4283
4
a X ~ N(6, 1²)
4.5 − 6
1
= P(Z < −1.5)
172 − 177
P(X < 172) = P Z <
7
= P(Z < −0.714)
)
= 0.44
5
= 0.0389
3
) (
11.2 − 10.8
9.2 − 10.8
−P Z <
1.6
1.6
)
= 1 − 0.8944
= 0.1056
b P(55 < X < 65) = P(X < 65) − P(X < 55)
(
) (
65 − 60
55 − 60
−P Z <
4
4
X ~N(10.8, 1.6 2)
=P Z <
15.5 − 10.8  = P Z > 2.938

a P (X > 15.5) = P  Z >
(
)


1.6
= P(Z < 1.25) − P(Z < −1.25)
)
= Φ(1.25) − Φ(−1.25)
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4 Normal distribution
= Φ(1.25) − [1 − Φ(1.25)]
= 1 − Φ(1.7)
= Φ(1.25) + Φ(1.25) − 1
= 1 − 0.9554
= 0.8944 + 0.8944 − 1
= 0.0446
= 0.7888
c P(X > 65) = 1 − P(X < 65)
65 − 60
=1−P Z <
4
(
c P(X > 85) = 1 − P(X < 85)
= 1 − P  Z < 85 − 68.5 


5
)
= 1 − P(Z < 3.3)
= 1 − P(Z < 1.25)
= 1 − Φ(3.3)
= 1 − Φ(1.25)
= 1 − 0.9995
= 1 − 0.8944
= 0.0005 = 0.05%
= 0.1056
9
d P(65 < X < 68) = P(X < 68) − P(X < 65)
(
) (
68 − 60
65 − 60
=P Z <
−P Z <
4
4
2.05 − 2.025  = P Z < 1.667

a P( X < 2.05) = P  Z <
(
)

0.015 
)
= Φ(1.667 ) = 0.9515 + 0.0007 = 0.9522
= P(Z < 2) − P(Z < 1.25)
b P( X > 2) = P  Z > 2 − 2.025 

0.015 
= Φ(2) − Φ(1.25)
= 0.9772 − 0.8944
= 0.0828
7
= P( Z > −1.667 ) = Φ(1.667 ) = 0.9522
10 a X ~ N (250, σ 2)
X ~ N (168, 4.62)
158 − 168  = P Z < −2.174

a P( X < 158 ) = P  Z <
(
)

4.6 
P (X < 235) = 0.005
=1– Φ (2.174) = 1 – (0.9850 + 0.0002) = 0.0148
z = – 2.576
235 − 250 = −2.576
σ
σ = 5.82
b P(165 < X < 170 ) = P( X < 170 ) − P( X < 165)
68 
= P  Z < 170 − 168  − P  Z < 165 − 16


4.6 
4.6 
= Φ (0.435) – (1 – Φ (0.652))
= (0.664+ 0.0018) – (1 – (0.7422 + 0. 0007))
= 0.4087
Φ(z) = 0.005
b P(X > 260) = 1 – P(X < 260) = 1 – Φ(1.717)
= 1 – 0.9570 = 0.0430
= P (Z < 0.435) – P (Z < – 0.652)
8
X ~ N(2.025, 0.0152)
Exercise 4.3B
1
a z = Φ−1(0.7642) = 0.720
About 40.9% of the students meet George’s
criteria.
b z = Φ−1(0.638) = 0.353
a X ~ N(68.5, 5²)
P(X > 70) = 1 − P(X < 70)
70 − 68.5
=1−P Z <
5
(
)
= 1 − P(Z < 0.3)
d P(Z < z) = 1− 0.375
z = Φ−1(0.625) = 0.318
e P(Z > z) = P(Z < −z) = 0.8531
= 1 − 0.6179
z = −1.05
f P(Z > z) = P(Z < −z) = 0.6
= 0.3281
−z = Φ−1(0.6) = 0.253
b 10% increase of 70 km/h gives 77 km/h
P(X > 77) = 1 − P(X < 77)
(
z = Φ−1(0.805) = 0.859
−z = Φ−1(0.8531) = 1.05
= 1 − Φ(0.3)
=1−P Z <
c P(Z < z) = 1 − 0.195
77 − 68.5
5
= 1 − P(Z < 1.7)
)
z = −0.253
g P(Z < z) = 0.372 so P(Z < −z) = 1 − 0.372
−z = Φ−1(0.628) = 0.327
z = −0.327
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4
WORKED SOLUTIONS
b 0.9798 = Φ(z)
h P(Z < 0) + P(0 < Z < z) = 0.5 + 0.45
z = 2.05 = x − 30
10
x = 50.5
0.45
0.5
c If P(X > x) = 0.1949, P(X < x) = 0.8051
0.8051 = Φ(z)
0
z = 0.86 = x − 30
10
x = 38.6
z
P(Z < z) = 0.95
d 0.75 = Φ(z)
z = Φ−1(0.95) = 1.645
z = 0.674 = x − 30
10
x = 36.74
i P(Z < −z) + P(−z < Z < z) = 0.4 + 0.5
z = Φ−1(0.9) = 1.282 (using the critical value for
the normal distribution table) or 1.281
0.4
e
P(X < x) = 0.35
P(Z < −z) = 0.65
0.4
Φ−1(0.65) = −z
z = −0.385 = x − 30
10
x = 26.15
0.1
f If P(X > x) = 0.05, P(X < x) = 0.95
0.95 = Φ(z)
−z 0 z
2
z = −1.036 =
3
z = 1.645 =
Underweight: 0.15 = Φ(z)
x − 498.7
7.3
x = 491.14
Therefore, bags of sugar are rejected when the
weight is less than 491.1 g.
Overweight: 0.9 = Φ(z)
z = 1.282 = x − 498.7
7.3
x = 508.06
Therefore, bags of sugar are rejected when the
weight is more than 508.1 g.
a X ~ N(19, 2.4²)
P(X > a) = 0.432, therefore,
P(X < a) = 1 − 0.432 = 0.568
x = 46.45
5
X ~ N(4000, 375²), P(X < x) = 0.9
0.9 = Φ(z)
z = 1.282 = x − 4000
375
x = 4480 hours
6
X ~ N(1.753, 0.1²), P(X < x) = 0.929
0.929 = Φ(z)
z = 1.468 = x − 1.753
0.1
x = 1.90 m
7
X ~ N(68.5, 5²), P(a < X < b) = 0.8
0.8 = P(X < b) − P(X > a)
However, since the normal distribution is
symmetrical about the mean,
0.9 = P(X < b), such that a = m − (b − m)
Therefore, 0.9 = Φ(z)
0.568 = Φ(z)
z = 0.171= a − 19
2.4
a = 19.41
b
z = 1.282 = b − 68.5
5
b = 74.91, a = 62.09
The range of speeds is between 62.1 mph and
74.9 mph.
P(X < b) = 0.205
P(Z > −z) = 0.205
1 − 0.205 = Φ(−z)
Φ−1(0.795) = −z
z = −0.824 = b − 19
2.4
b = 17.02
4
a 0.99 = Φ(z)
z = 2.326 =
x = 53.26
x − 30
10
x − 30
10
8
a Marks scored by students are independent.
b X ~ N(80, 12²), P(X > x) = 0.25, P(X < x) = 0.75
0.75 = Φ(z)
z = 0.674 = x − 80
12
x = 88.088
Therefore, the lowest mark needed is 89.
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4 Normal distribution
b $45 ÷ $1.55 = 29.03 litres
c P(a < X < b) = 0.95, and since the normal
distribution is symmetrical, P(X < b) = 0.975
(
P ( X > 29.03) = P Z > 29.03 − 35
8
0.975 = Φ(z)
= P ( Z > −0.746 ) = Φ ( 0.746 ) = 0.7722
z = 1.96 = x − 80
12
b = 103.52, a = 56.48
c Lorries may also fill up at the petrol station,
which will affect the normal distribution
parameters, as will the amount of fuel
remaining in the petrol station.
Therefore, the middle 95% of marks will be
between 56 and 104.
d P(X > 90) = 1 − P(X < 90)
(
90 − 80
=1−P Z <
12
Some people may just visit the counter at the
petrol station rather than purchase petrol.
This will also affect the normal distribution
parameters.
)
= 1 − P(Z < 0.833)
= 1 − Φ(0.833)
= 1 − 0.7975
Exercise 4.3C
1
= 0.2025
200 × 0.2025 = 40.5
Therefore, 40 students are likely to score 90
or more.
9
X ~ N(190, 2.52)
2
aP (X > x) = 0.2
1 – Φ(z) = 0.2
Φ(z) = 0.8
z = 0.842
190
x
−
= 0.842
2.5
3
x = 192.1 g
b P( –z < Z < z) = 0.4
Φ(z) – (1 – Φ(z)) = 0.4
Φ(z) = 0.7
z = 0.524
x − 190
= 0.524
2.5
)
x − 190
= −0.524
2.5
x = 191.3, 188.7
Therefore, the range in weights is from 188.7 g to
191.3 g.
c P( X 195) = P  Z 195 − 190  = P( Z 2)

2.5 
= 1 − Φ( 2) = 1 − 0.9772 = 0.0228
100 × 0.0228 = 2.28
At least two tins are likely to be 195 g or more.
10 X ~ N(35, 82)
20 − 35  = P Z < −1.875

a P( X < 20 ) = P  Z <
(
)

8 
= 1– Φ(1.875) = 1 – (0.9693 + 0.0004) = 0.0303
4
X ~ N(43, σ ²)
P(X > 48) = 0.2
Therefore, P(X < 48) = 0.8
0.8 = Φ(z)
z = 0.842 = 48 − 43
s
σ = 5.94
X ~ N( m, 1.3²)
P(X > 47) = 0.11
Therefore, P(X < 47) = 0.89
0.89 = Φ(z)
47 − µ
z = 1.227 =
1.3
m = 45.4 cm
X ~ N(m, σ ²)
P(X > 332.91) = 0.18, P(X > 325.42) = 0.72
Therefore, P(X < 332.91) = 0.82
332.91 − m
= Φ(0.82)
s
332.91 − m
= 0.915
s
332.91 − m = 0.915σ
Also P(X < 325.42) = 0.28
325.42 − m
= Φ(0.28)
s
325.42 − m
= −0.583
s
325.42 − m = −0.583σ
1 − 2:
7.49 = 1.498σ
σ = 5 ml
Therefore, m = 325.42 + 0.583(5) = 328 ml
1
2
X ~ N(m, σ ²)
P(X < 9.92) = 0.25, P(X > 12.24) = 0.25
P(X < 9.92) = 0.25
9.92 − m = Φ(0.25)
s
9.92 − m
= −0.674
s
9.92 − m = −0.674σ
1
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WORKED SOLUTIONS
Also P(X < 12.24) = 0.75.
12.24 − m = Φ(0.75)
s
12
.
24
− m = 0.674
s
12.24 − m = 0.674σ
2 − 1:
2.32 = 1.348σ
σ = 1.721
Therefore, m = 12.24 − 0.674(1.721) = 11.08
= 1 − Φ(1)
= 1 − 0.8413
= 0.1587
c P(87 < X < 95) = P(X < 95) − P(X < 87)
2
= 0.9522 − 0.1587
= 0.7935
d X ~ N(μ, σ 2)
P(X > 85) = 0.1
Alternative method to find the mean:
(
a
)
( σ µ)
Φ
X ~ N(15.5, σ 2)
85 −
Φ(z) = 0.001
z = 2.326
)
( σ µ)
σ = 1.07 cm
b P( X < 14 ) = P  Z < 14 − 15.5  = P( Z < −1.395)

1.075 
=1 − Φ(1.395) = 1 − (0.9177 + 0.0008) = 0.0815
Φ −
65 −
−
8.15% pebbles are smaller than 14 cm.
1
65 − µ
= 0.2
σ
P Z <
= 0.8
65 − µ
= 0.842
σ
–(65 – μ) = 0.842 σ X ~ N(500, σ 2)
2
1 + 2
a P(X < 480) = 0.1
20 = 2.123 σ
1 – Φ(z) = 0.90
– z = – 1.282
480 − 500
= −1.282
σ
σ = 15.6 g
(
b P(X > 508) = P Z >
7
= 0.9
85 – μ = 1.281 σ
(
18 − 15.5
= 2.326
σ
6
P(X < 65) = 0.2
85 − µ
= 1.281
σ
P(X >18) = 0.01
and
85 − µ
P Z >
= 0.1
σ
LQ = 9.92 and UQ = 12.24
Therefore, µ = 9.92 + 1 (12.24 – 9.92)
2
= 9.92 + 1.16 = 11.08
5
4
Therefore, μ = 85 – 1.281 × 9.42 = 72.9 minutes
8
)
508 − 500
= P ( Z > 0.506 )
15.60
= 1 – Φ(0.506)
= 1 – (0.6915 + 0.002) = 0.3065
a X ~ N(90, 3²)
P(X > 95) = 1 − P(X < 95)
95 − 90
=1−P Z <
3
(
= 1 − Φ(1.667)
= 1 − 0.9522
= 0.0478
(
87 − 90
3
= P(Z < −1)
b P(X < 87) = P Z <
= Φ(−1)
σ = 9.42 minutes
)
)
X ~ N( m, σ ²)
P(X ⩾ 140) = 0.2, P(X ⩾ 130) = 0.8
Therefore, P(X < 140) = 0.8
140 − m = Φ−1(0.8)
s
140 − m = 0.842
s
140 − m = 0.842 σ
Also P(X > 130) = 0.8
P(Z > z) = 0.8
P(Z < −z) = 0.8
130 − m
= −Φ−1(0.8)
s
130 − m
= −0.842
s
130 − m = −0.842σ
1 − 2:
10 = 1.684σ
σ = 5.94 kg
Therefore, m = 140 − 0.842(5.94) = 135 kg
1
2
Kk
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4 Normal distribution
9
X ~ N(12, 22)
Exercise 4.4A
a P( X < 11) = P  Z < 11 − 12 

2 
1
a m = np = 20 × 0.5 = 10
b Variance: σ 2 = np(1 − p) = 20 × 0.5 × 0.5 = 5
= P( Z < −0.5) = 1 − Φ( 0.5)
Standard deviation: σ = 5 = 2.236
= 1 − 0.6915 = 0.3085
(
13.5 − 12
b P ( X > 13.5) = P Z >
2
)
c P(15.5 < X < 16.5) = P
(15.52.236− 10 < Z < 16.52.236− 10 )
= P(2.460 < Z < 2.907)
= P ( Z > 0.75) = 1 − Φ ( 0.75)
= Φ(2.907) – Φ(2.460)
= 1 − 0.7734 = 0.2266
= 0.9982 − 0.9931
( X) =<P10 )Z+<P10
( X −>1213) = P Z < 10 −2 12 
c P( X < 10 ) + P( X >P13
= 0.0051

2 
10−13
12− 12 
12
10
−


10
13
P
X
<
+
P
X
>
=
P
Z
<
) ) =+( PP13

−)12  
P( X < 10 ) +( P( X > 13
Z<
5) < −1) + P( Z > 0.5) 2 a µ = 108 × 0.88 = 95.04
 2 2= P( Z+P<2−1) 2+ P(Z=>P0(.Z
σ2 = 108 × 0.88 × 0.12 = 11.4048
 13 − 12  = P Z < −1 + P Z > 0.5
−
13
12


+
P
) .5) (
)
= P Z <−1 +( P Z > 0
σ = 3.377
+P 
 2   (2  ) (
99.5 − 95.04
P(X < 99.5) = P Z <
= (1 – Φ (1)) + (1 – Φ (0.5)) = (1 – 0.8413) + 0.3085
3.377
= Φ(1.321)
= 0.4672
(
= 0.9067
d P(X > 13) = 0.05 and P(X < 10) = 0.12
b P(85.5 < X < 104.5)
1 – Φ(z1) = 0.95 and 1 – Φ(–z2) = 0.88
(
= P 85.5 − 95.04 < Z < 104.5 − 95.04
3.377
3.377
z1 = 1.645 –z2 = 0.1171
13 − µ
= 1.645
σ
13 – µ = 1.645σ
10 − µ
= −1.171
σ
–10 + µ = 1.171σ
)
)
= P(Z < 2.801) − P(Z < −2.825)
= Φ(2.801) – (1 − Φ(2.825))
1
= 0.9975 − (1 − 0.9976)
= 0.9951
2
3
1 − 2:
3 = 2.816σ
m = np = 200 × 0.32 = 64
Variance: σ 2 = np(1 − p) = 200 × 0.32 × 0.68 = 43.52
Standard deviation: σ = 43.52 = 6.597
129.5 − 64
a P(X < 129.5) = P Z < 6.597
σ = 1.07 minutes
(
Therefore, µ = 13 – 1.645 × 1.07 = 11.2 minutes
)
= Φ(9.93)
12 − 16
P ( X < 12) + P12
18 ) = P Z <
( X −>16
≈1
4.5
a P ( X < 12) + P ( X > 18 ) = P Z <
4.5
12
−
16
b
P(X
>
50.5)
=
P  Z > 50.5 − 64  = P(Z > − 2.046)
18
−
16
>
) +)P=( PX 18
)12= −P16Z <+ P4.5

6.597 
= P ( Z < −0.889) + P ( Z > 0.444 )
> 18
Z 18
<
( X < 12) P+(PX( X< 12
− 16
4.5 ) + P ( Z > 0.444 )
4.5
+P
= P ( Z < −0.889
4.5
=
P(
Z
<
2.046)
18 − 16+ P 18 − 16 = P ( Z < −0.889) + P ( Z > 0.444 )
P
= P (4.5
Z < −0.889) + P ( Z > 0.444 )
4.5
= Φ(2.046)
= (1 – Φ (0.889)) + (1 – Φ (0.444))
= 0.9796
= (1 – (0.8106 + 0.0025)) + (1 – (0.67 + 0.0014))
75.5 − 64
c P(X > 75) = P Z >
6.597
= 0.5155
= 1 − Φ(1.743)
b P(X < x) = 0.9, Φ(z) = 0.9, z = 1.282
10 X ~ N(16, 4.52)
(
)
)(
((
(
)() ( ) )
)
) (
(
x − 16
= 1.282
4.5
= 1 − 0.9594
x = 21.769, so 22 minutes
= 0.0406
d P(75 < X < 130) = P(X < 129.5) − P(X < 75.5)
c X ~ N(16, σ 2)
P(X < 19) = 0.99, Φ(z) = 0.99, z = 2.326
19 − 16
= 2.326
σ
σ = 1.29 minutes
)
= 1 − 0.9594
4
= 0.0406
a
You may use the normal distribution as an
approximation for the binomial B(n, p) (where n
is the number of trials each having probability p
of success) when:
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4
WORKED SOLUTIONS
b P ( 44.5 X 45.5) = P ( X 45.5) − P ( X 44.5)
n is large
(
This ensures that the distribution is
reasonably symmetrical and not skewed at
either end.
= Φ(0.972) – Φ(0.795)
= (0.8340 + 0.0005) – (0.7852 + 0.0015) = 0.0478
c Mean = m = 15 × 0.4 = 6
8 a X ~ B (30, 0.12)
Variance = σ 2 = np(1 − p) = 15 × 0.4 × 0.6 = 3.6
(
4.5 − 6
1.897
= P(Z < −0.791)
P(T < 5) = P Z <
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
 30
 30
0
30
1
29
=   ( 0.12) ( 0.88 ) +   ( 0.12) ( 0.88 )
0
1 
 30
 30
=   30
(0.12
)0 (0.88
)30 +28  (0.12)1 (0.88 )29

2
 0+ 
1 
 2  ( 0.12) ( 0.88 )
 30
2
28
+   ( 0.12) ( 0.88 )
2
np(1 − p) = 1.897
)
= 1 − Φ(0.791)
= 1 − 0.7855
= 0.2847
= 0.2145
d
5
(0.2173 − 0.2145) × 100% = 1.29%
b µ = np = 60 × 0.12 = 7.2
0.2173
np (1 – p) = 60 × 0.12 × 0.88 = 6.336
n = 120, p = 0.5
σ = 6.336 = 2.517
µ = np = 120 × 0.5 = 60
P ( X < 8.5) = P Z <
(
np(1 – p) = 120 × 0.5 × 0.5 = 30
σ = 30 = 5.477
(
P ( X 54.5) = P Z )
54.5 − 60
= P ( Z −1.004 )
5.477
a X ~ B(50, 0.35)
 50 
20
30
P ( X = 20 ) =   ( 0.35) ( 0.65) = 0.0875
 20
b µ = np = 50 × 0.35 = 17.5
σ = 11.375 = 3.373
X ~ N(17.5, 11.375) P (19.5 X 20.5)
X
~ (N(17.5,
11.375)
19.5
) X 20.5)
X ~ N(17.5, 11.375) P
P
X
X
20.5
19.5
=(19.5
− P ()XP (
)20.5
) (
)) (
)
)
µ = np = 200 × 0.2 = 40
np (1 – p) = 200 × 0.2 × 0.8 = 32
σ = 32 = 5.657
Approximate X ~ B(200, 0.2) with X ~ N(40, 32)
)
28.5 − 40
a P ( X 29) = P Z > 5.657 = P ( Z > −2.033)
= Φ ( 2.033) = 0.9790
15
0 15
14
1
15 ×
0.72 × 0.28 + C14 × 0.72 × 0.28
15
13
2
+ C13 × 0.72 × 0.28
So: P(at least 13) = 0.1645
b Mean = m = 15 × 0.72 = 10.8
Variance = σ 2 = np(1 − p) = 15 × 0.72 × 0.28 = 3.024
(
= (0.8106 + 0.0025) – (0.7224 + 0.0010) = 0.0897
(
15C
=1−P Z <
= P(Z  0.889) – P (Z  0.593)
= Φ(0.889) – Φ(0.593)
7
a To find the probability of at least 13:
np(1 − p) = 1.739
P(X  13) = 1 − P(X < 12.5)
P (19.5
X ) 20.5
P ( X 19.5) 19.5 − 17.5
= P ( X 20.5) − P ( X=
20.5) −−17.5
=P Z
−P Z 3.373
3.373
20.5
−
17.5
− 17.5
− 17.5 − P Z 19.5
= P Z 20.5 − 17.5 =−PPZZ 19.5
3.373
3.373
3.373
3.373
(
9
Standard deviation =
np (1 – p) = 50 × 0.35 × 0.65 = 11.375
(
) ((
)
8.5 − 7.2
= P ( Z < 0.516 )
2.517
= Φ ( 0.516 ) = 0.6971
= Φ(1.004) = 0.8413 + 0.0009 = 0.8422
6
)
= P(Z , 0.972) – P(Z , 0.795)
b P(T < 5) = 0.2173
Standard deviation =
) (
= P Z 45.5 − 40 − P Z 44.5 − 40
5.657
5.657
np > 5 and n(1 – p) > 5
12.5 − 10.8
1.739
)
= 1 − Φ(0.978)
= 1 − 0.8360
= 0.1640
c (0.1645 − 0.1640) × 100% = 0.3% error, so the two
0.1645
values agree closely.
d The agreement would be worse, as the normal
distribution is a better approximation when
p is closer to 0.5. When p = 0.85, n(1 − p) = 2.25,
which is smaller and further from 5 than its
value when p = 0.72 (4.2).
1
10 a P(any one correct) =
5
b Mean = m = 35 × 0.2 = 7
Variance = σ 2 = np(1 − p) = 35 × 0.2 × 0.8 = 5.6
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4 Normal distribution
= 1 − Φ (1.333)
c P(X < a) = 0.99
Φ−1(0.99) = z
z = 2.326
a−7
2.326 =
5.6
a – 7 = 2.326 × 5.6
a = 12.504
So the pass mark should be 13.
= 0.0913
0.0913 × 12 = 1.09 months
b P(Y < 53) = P Z < 53 − µ  = 0.38

5 
53 − µ
Φ−1(0.38) =
5
53 − µ
5
−1.525 = 53 − m
−0.305 =
Exam-style questions
1
a X ~ N(1005, 2²)
(
1000 − 1005
2
= P(Z < −2.5)
= Φ(−2.5)
= 1 − Φ(2.5)
= 1 − 0.9938
= 0.0062
P(X < 1000) = P Z <
)
b P(X > 1007) = 1 − P(X < 1007)
1007 − 1005
=1−P Z <
2
(
m = 54.5 mm
(
60 − 54.53
5
= Φ(1.094)
c P(Y < 60) = P Z <
= 0.863
 12
P(exactly 4 months) =   (0.863)4(0.137)8
 4
= 0. 000 034 1
)
5
a X ~ N(1000, 110²)
(
= 1 − 0.8413
900 − 1000
110
= P(Z > −0.909)
= 0.1587
= Φ(0.909)
P(X > 900) = P Z >
= 1 − Φ(1)
c P(1000 < X < 1007) = 1 − (0.1587 + 0.0062)
= 0.8351
So the probability that they both contain
between 1000 g and 1007 g is 0.83512 = 0.697.
2
(
a P(taller than 160 cm) = P Z > 160 − 164.5
8.75
= Φ(0.514)
= 0.6964
)
= 0.8182
(
)
= 1 − Φ(1.818)
= 0.0345
c P(900 < X < 1000) = P(X < 1000) − P(X < 900)
1000 − 1000
= P Z <
110
(
(
−P X <
P(Z < z) = 0.45
)
900 − 1000
110
)
= P(Z < 0) − P(Z < −0.909)
P(Z < −z) = 0.55
Φ−1(0.55) = −z
= Φ(0) − Φ(−0.909)
= Φ(0) − [1 − Φ(0.909)]
z = −0.125 = h − 164.5
8.75
h = 163.4 cm
P(X > 320) = 1 – P(X < 320)
 320 − 4( 72) 

=1−Φ 
 4 × 10 2 


= 1 − 0.9452
= 0.0548
(
4
)
b P(X > 1200 hours) = P Z > 1200 − 1000
110
= P(Z > 1.818)
bP(H < h) = 0.45
3
)
a X ~ N(82, 62)
(
P(X > 90) = P Z >
90 − 82
6
)
)
= 0.5 − 1 + 0.8182
= 0.3182
dP(X < x) = 0.01
P(Z < z) = 0.01
P(Z < −z) = 0.99
−z = 2.326
x − 1000
= −2.326
110
x = 744.14 hours
or 31 days to the nearest day.
52
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WORKED SOLUTIONS
6
b X ~ N(m, 0.13²)
a X ~ B(30, 0.2)
P(X < 3.54) = 0.04
Approximated by Y ~ N(6, 4.8) since
np = 6 > 5 and nq = 24 > 5.
3.54 − µ 
= P Z <

0.13 
b P( X 4) = P(Y > 3.5)
3.5 − 6 

= P Z >


4.8 
z = −Φ−1(0.96)
 3.54 − µ  = −1.751
 0.13 
3.54 − m = −1.751× 0.13
= P( Z > −1.141)
m = 3.77 litres
= Φ(1.141) = 0.8731
c P(X 4 on each of two days)
c X ~ N(3.58, σ ²)
= 0.8731 × 0.8731 = 0.7623
7
P(X < 3.54) = 0.04
a X ~ N(m, σ ²)
P(X > 212.6) = 0.33
Therefore, 0.33 = 1 − P(X < 212.6)
= 1 − P  Z < 212.6 − µ 


σ
212.6
−
µ
Φ−1(0.67) =
σ
212.6 − µ
0.440 =
σ
212.6 − m = 0.44σ
1
Also 0.12 = P(X < 211.8)
P(Z < z) = 0.12 is the same as P(Z < −z) = 0.88.
Φ−1(0.88) = −  211.8 − µ 


σ
−211.8 + m = 1.175σ
1 + 2:
0.8 = 1.615σ
σ = 0.495
Therefore, m = 212.6 – 0.44(0.495) = 212.38
P(X > 212) = P  Z > 212 − 212.38 

0.4954 
= P(Z > −0.767)
= 0.8017 − 0.12
= 0.682
Therefore, 31.8% are rejected because they are
outside this range.
a X ~ N(3.58, 0.13²)
(
3.54 − 3.58
0.13
= P(Z < −0.308)
P(X < 3.54) = P Z <
= Φ(−0.308)
= 1 − Φ(0.308)
(
3.54 − 3.58
s
z = −Φ−1(0.96)
=P Z <
−
9
)
0.04 = −1.751
s
σ = 0.0228 litres
X ~ N(416, 8σ ²)
P(X > 400) = 0.99
Therefore, P Z > 400 − 416 = 0.99
2.828s
(
(
P Z<
)
−16
2.828s
) = 0.01
−16 = −2.326
2.828s
σ = 2.432
2
So the standard deviation is 2.43 g.
10 aThe number of trials is large, making the
binomial distribution impractical as a model.
b np = 2500 × 0.38 = 950 and nq = 2500 × 0.62 = 1550
Since np  5 and nq  5, a normal
approximation can be used.
c m = np = 950 and σ 2 = npq = 589, so X ~ N(950, 589)
= P(Z < 0.767)
= 0.7785
b P(accepted) = P(211.8 < X < 212.8)
= P(X < 212.8) − P(X < 211.8)
= P  Z < 212.8 − 212.38  − 0.12


0.4954
8
4
)
d 40% of the vote is 1000 votes.


P( X > 1000 ) = P  Z > 1000 − 950 

589 
= P(Z > 2.06)
= 1 − P(Z < 2.06)
= 1 − 0.9803
= 0.0197
11 X ~ N(75, 122)
a From the tables: P(Z < 1.281) = 0.9
x − 75
Therefore, 1.281 =
⇒ x = 90.372
12
Assuming that only whole marks are
awarded, students must score at least 91
marks to be in the top 10%.
= 1 − 0.6209
= 0.3791
53
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4 Normal distribution
b From the tables: P(z < 0.674) = 0.75
Therefore, 0.674 = x − 75 ⇒ x = 83.1
12
By symmetry, the middle 50% of the
students score between 67 marks and 83
marks (to nearest mark).
(
c P(X < 40) = P Z < 40 − 75
12
From the tables: P(z < 1.724) = 0.9577
Therefore, 1.724 = 15 − 8.75 ⇒ σ = 3.63 minutes
s
b P(X < 5) = P z < 5 − 8.75 

3.625 
)
= P(z < −1.034)
= 1 − P(z < 1.034)
= 1 − 0.8494
= P(Z < −2.917)
= 0.1506
= 1 − P(Z < 2.917)
c 15 000 × 0.1506 = 2259 patients
= 1 − 0.9982
14 M ~ B(200, 0.44)
= 0.0018
np = 200 × 0.44 = 88
12 X ~ N(55, 172)
a Equal proportions means equal probabilities
42 − 55
P(X < 42) = P Z < 17
= P(Z < −0.765)
(
)
nq = 200 × 0.56 = 112
Since 88 > 5 and 112 > 5, a normal approximation
can be applied.
np(1 – p) = 200 × 0.44 × 0.56 = 49.28
= 1 − P(Z < 0.765)
M ~ N (88, 49.28)
= 1 − 0.7779
P ( 75.5 < M < 99.5) = P ( M 99.5) − P ( M 75.5)
= 0.2221
The 0.7779 remaining for standard eggs and
large eggs needs to be divided equally. This
means that P(X < x) = 1 − 0.7779 = 0.6111
2
From the tables: P(z < 0.282) = 0.6111
Therefore, 0.282 = x − 55 ⇒ x = 59.79
17
So eggs weighing more than 59.8 grams
should be considered to be large eggs.
(
)
b X ~ B(1000, 0.01)
np = 1000 × 0.01 = 10 and nq = 1000 × 0.99 = 990.
Since np ≥ 5 and nq ≥ 5, a normal
approximation can be used.
X ~ N(10, 9.9)
To calculate P(X = 10), we need to use a
continuity correction.
P(X = 10) ≈ P(9.5 < X < 10.5)
≈ P(X < 10.5) − P(X < 9.5)


10.5 − 10 
9.5 − 10 
≈ PZ <
 − P  Z <


9.9 
9.9 
≈ P(Z < 0.159) − P(Z < −0.159)
≈ P(Z < 0.159) − [1 − P(Z < 0.159)]
≈ 2P(Z < 0.159) − 1
≈ 2 × 0.5632 − 1




= P  Z 99.5 − 88  − P  Z 75.5 − 88 


49.28 
49.28 
= P(Z , 1.638) – P (Z , – 1.781)
= Φ (1.638) – (1 – Φ (1.781))
= (0.9484 + 0.0008) – 1 + (0.9625 + 0.0001) = 0.912
15 a X ~ B(12 000, 0.003)
b μ = np = 12 000 × 0.003 = 36
σ 2 = np (1 – p) = 12000 × 0.003 × 0.997 = 35.892
c np = 36
nq = 12 000 × 0.997 = 11 964
Since 36 > 5 and 11964 > 5, a normal
approximation can be applied.
Y ~ N (36, 35.892)
P(29.5 Y 30.5) = P(Y 30.5) − P(Y 29.5)

30.5 − 36  P  Z 29.5 − 36 
= P Z  −  

35.892 
35.892 
= P(Z −0.918) − P(Z −1.085)
= (1 − Φ(0.918)) − (1 − Φ(1.085))
= Φ(1.085) − Φ(0.918)
= (0.8599 + 0.0012) − (0.8186 + 0.002) = 0.0405
16 T ~ N(75, 152)
a i
a P(X < 15) = 0.9577
)
60 − 75
= P ( Z < −1)
15
= 1 − Φ (1) = 1 − 0.8413 = 0.1587
= 0.1264
13 P(X > 15) = 0.0423 and X ~ N(8.75, σ 2)
(
P (T < 60 ) = P Z <
ii P ( 60 T 90 ) = P (T 90 ) − P (T 60 )
(
=P Z <
) (
90 − 75
60 − 75
−P Z <
15
15
)
54
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WORKED SOLUTIONS
= P(Z < 1) – P(Z < – 1) = Φ(1) – (1 – Φ(1))
= 0.8413 – 1 + 0.8413 = 0.6826
b Since the mean time a user spends in the
library is 75 minutes, which is more than an
hour, a user coming after 9.00 pm can only
spend under the mean time.
19 W ~ B (12, 0.48)
 12
4
8
a P ( X = 4 ) =   ( 0.48 ) ( 0.52) = 0.1405
 4
b 3 of a bag = 9 bulbs
4
P(X > 9) = P(X = 10) + P(X = 11) + P(X = 12)
c You know that 99.75% the data lies within ±3 σ.
 12 
10
2  12 
11
1
=
 ( 0.48 ) ( 0.52) +  11  ( 0.48 ) ( 0.52)
10




m = 75 σ = 15
m + 3σ = 75 + 3 × 15 = 120 minutes = 2 hours
 12 
12
+
 ( 0.48 ) = 0.0137
 12 
10 – 2 = 8
The latest time of entry to the library is
about 8 pm.
17 a n = 300, p = 0.02
c n = 50, p = 0.48
np = 50 × 0.48 = 24
np = 300 × 0.02 = 6
nq = 50 × 0.52 = 26
nq = 300 × 0.98 = 294
Since 24 > 5 and 26 > 5, a normal approximation
can be applied.
Since 6 > 5 and 294 > 5, a normal approximation
can be applied.
np(1 – p) = 50 × 0.48 × 0.52 = 12.48
np (1 – p) = 300 × 0.04 × 0.98 = 5.88
X ~ N (24, 12.48)
X ~ N (6, 5.88)


P( X > 30.5) = P  Z > 30.5 − 24 

12.48 


P( X < 4.5) = P  Z < 4.5 − 6 

5.88 
= P( Z > 1.840 ) = 1 − Φ(1.840 )
= P( Z < −0.619) = 1 − Φ( 0.619)
= 1 – (0.7291 + 0.0029) = 0.268
b The cost to produce 300 dolls: 1.50 × 300 = $450
= 1 – 0.9671 = 0.0329
20 X ~ B (n, 0.07)
a E(X) = 8
The expected number of damaged dolls
= 300 × 0.02 = 6
np = n × 0.07 = 8 n = 114.25 Therefore, 115 people
The revenue from the dolls = (300 – 6) × 10 = $2940
b np = n × 0.07 = 2 n = 29
Therefore, the expected profit = $2940 – $450
= $2490
18 a W ~ N (500, 152)
P(W < w) = 0.1
σ 2 = np(1 – p) = 29 × 0.07 × 0.93 = 1.8879
Therefore, σ = 1.8879 = 1.374
Mathematics in life and work
Φ (–z) = 0.9
1
z = − 1.282
w − 500 = −1.282
15
w = 480.8
b P ( 505 W 515) = P (W 515) − P (W 505)
(
4
) (
P(X > 308 g) = 1 − P(X < 308)
308 − 300
=1−Φ
11
(
)
= 1 − 0.7663
= 0.2337
)
2
Find m such that P(X < m) = 0.02
P(Z < z) = 0.02
P(Z < −z) = 0.98
= P(Z < 1) – P(Z < 0.333) = Φ(1) – Φ(0.333)
−z = Φ−1(0.98)
= 0.8413 – (0.6293 + 0.0011) = 0.2109
z = −2.054
Let Y be the number of Chinese leaf cabbages
m − 300 = −2.054
that Amy picked, Y ~ B (5, 0.211)
11
 5
m = 277.4 = 277 g (3 s.f.)
3
2  5
4
1  5
5
P (Y 3) =   ( 0.211) ( 0.789) +   ( 0.211) ( 0.789) +   ( 0.211) = 0.0667
3 5Y ~ N(m, σ ²),
 3
 4
P(Y > 290) = 0.95, P(Y < 305) = 0.97
 5
3
2  5
4
1  5
5
290 − m
P (Y 3) =   ( 0.211) ( 0.789) +   ( 0.211) ( 0.789) +   ( 0.211) = 0.0667
Therefore, Φ−1(0.95) = −1.645 =
s
 3
 4
 5
=P Z <
515 − 500
505 − 500
−P Z <
15
15
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4 Normal distribution
290 − m
s
290 − m = −1.645 σ
305 − m
Also, Φ−1 (0.97) = 1.881 =
s
305 − m
1.881 =
s
305 − m = 1.881σ
2 − 1:
15 = 3.526 σ
σ = 4.25 g
Therefore, m = 305 − 1.881(4.25) = 297 g
−1.645 =
1
2
56
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WORKED SOLUTIONS
Summary Review
Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the
question. In some cases, alternative methods are shown for contrast.
All sample answers have been written by the authors. Cambridge Assessment International Education bears no
responsibility for the example answers to questions taken from its past question papers, which are contained in this
publication.
Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in
degrees, unless a different level of accuracy is specified in the question.
Use graph paper when drawing graphs; graphical solutions given here are not to exact scale but give an indication
of the answer.
Warm-up Questions
1
iii Drawing a line from 2200 hours shows:
i 0.15 + 0.22 + 0.18 + 0.24 = 0.79
200
Cumulative
frequency
The sum of the probabilities must sum to 1,
so missing value is 1 – 0.79 = 0.21.
ii P(lemon or lime) = 0.15 + 0.22 = 0.37
2
Paul:
9
= 0.257…
35
Sammy:
26% = 0.26
150
100
0
1000 1500 2000 2500 3000 3500
Lifetime (hours)
c P(exactly one bulb > 3500 hours) =
Lifetime, T hours
T  1000
T  1500
(
)(
Number of bulbs
1 3
9 2
P(exactly one bulb > 3500 hours) =
×
+
×
10 5
10 5
10
18 3
21
=
+
=
40
50 50 50
T  2000
95
T  2500
167
A Level Questions
T  3500
200
1
)
(109 × 52 ) + (101 × 35 )
X ~ N(μ, 0.7142) and the area we are interested
in is:
ii Be sure to plot cumulative frequency
against the upper class boundaries.
200
T
NB – this estimate could be as low as
68 depending on the quality of the diagram.
μ ≈ 1999 hours
Cumulative
frequency
500
Therefore, the number of bulbs lasting more
than 2200 hours is 200 – 120 = 80 bulbs.
µ ≈ 399 750
200
0.475
×
×
150
100
×
3.2
50
×
0
×
×
a µ ≈ ∑ fx = [(500 × 10) + (1250 × 30) + (1750 × 55)
∑ f + (2250 × 72) + (3000 × 33)] ÷ 200
b i
×
50
Therefore, Sammy is more likely to win because he
has the higher probability.
3
×
×
×
×
500
1000 1500 2000 2500 3000 3500
Lifetime (hours)
T
μ
Using the cumulative normal distribution tables
in reverse, we can say that P(Z < 1.96) = 0.975.
Therefore:
57
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Summary REVIEW
−1.96 =
3.2 − µ
0.714
4
iFor type A, there are 61 smartphones, so the
median is the 31st item in the ordered list.
−1.3994 … = 3.2 − μ
The median is 0.52 seconds.
μ = 4.5994 … = 4.60 m (3 s.f.)
2
The lower quartile is between the 15th and
16th items in the ordered list.
X ~ B(11, 0.76)
The lower quartile is:
0.40 + 0.42 = 0.41 seconds
2
The upper quartile is between the 15th and
16th items from the end of the ordered list.
P(X < 10) = 1 − [P(X = 10) + P(X = 11)]
 11

 11
= 1 −   × 0.7610 × 0.24 +   × 0.7611 
 11
 10

 11

 11
= 1 −   × 0.7610 × 0.24 +   × 0.7611 
11
10






= 0.781
3
The upper quartile is:
0.78 + 0.80 = 0.79 seconds
2
ii
iP(1st die divisible by 3 and 2nd die not divisible
2 4 2
by 3) = × =
6 6 9
But to find P(A), we need to consider that
this could happen the other way around, i.e.
with the 1st die not divisible by 3 and the
2nd die divisible by 3.
Therefore, P( A) = 2 × 2 = 4
9
9
We know that the product of two
odd numbers is an odd number.
So the probability of an odd product is:
3 3
9
1
× =
=
6 6 36 4
1 3
Therefore, P(B) = 1 − =
4 4
4 3 1
So P(A) × P(B) = × =
9 4 3
To find P(A ∩ B) it is useful to draw a sample
space diagram.
1
2
3
4
5
6
1
2
3
4
5
6
represents event A
represents event B
Yellow highlighting represents A ∩ B
Therefore, P(A ∩ B) = 12 = 1 .
36 3
Because P(A ∩ B) = P(A) × P(B), we can
conclude that A and B are independent.
ii Since P(A ∩ B) > 0, we can say that A and B
are not mutually exclusive.
Type A
Type B
0
0.2
0.4
0.6
0.8
1
Time (s)
iii The median for B is lower than for A, so
smartphone B is quicker because it takes
less time on average to load the apps. The
interquartile range for B is narrower too, so
smartphone B is more consistent.
5
a iIf we treat the 4 Es as though they are one
block, then we get (EEEE)***** and there are
6! arrangements of this list. However, there
are 2 Ns and 2 Ss, so we need to divide by 2!
twice to remove the duplicate arrangements.
The number of arrangements is 6! = 180 .
2!2!
iiSuppose that S is fixed in place at the
beginning of the list and that T is fixed
in place at the end of the list: S*******T.
There are 7! arrangements, but there
are 2 Ns and 4 Es, so we need to divide
by 4!2!. We get 7! = 105 . But we
4!2!
could repeat the process with S and T
reversed, which doubles the number of
arrangements to 210.
b There are 3 Es available. Assume that one
of them is selected and that the others are
then unavailable. We then need to choose 3
letters from the remaining 6 letters, which is
6C = 20.
3
There are 20 possible selections that contain
exactly one E.
58
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WORKED Solutions
6
i
1st
Attempt
iii The lower quartile is approximately the 28th
person, so in class 6–20.
3rd
Attempt
2nd
Attempt
The upper quartile is approximately the 83rd
person, so in class 61–80.
S
0.4
IQRmin = UQmin − LQmax
U
0.6
IQRmin = 61 − 20 = 41
S
0.4
S = P (successful)
U = P (unsuccessful)
8
S
0.4
U
0.6
µ = 5 − 2 + 12 + 7 − 3 + 2 − 6 + 4 + 0 + 8 = 2.7
10
Var(X) = E(X2) − E2(X)
Var( X ) = [52 + (−2)2 + 122 + 7 2 + (−3)2 + 22 + (−6)2 + 4 2 + 0 2 + 8 2] ÷ 10
U
2
Var( X ) = [52 + (−2)2 + 0.6
122 + 7 2 + (−3)2 + 2
+ (−6)2 + 4 2 + 0 2 + 8 2] ÷ 10 − 2.7 2
ii P(X = 0) = 0.4
Var(X) = 35.1 − 2.72 = 27.8
P(X = 2) = 0.62 × 0.4 = 0.144
9
P(X = 3) = 0.63 = 0.216
ii Firstly, find the class widths, then each
frequency density as shown in the table below.
So the completed table is:
0
1
2
3
P(X = x)
0.4
0.24
0.144
0.216
Length
(cm)
E(X) = 1.176
iFirstly, find the class widths, then each
frequency density as shown in the table below.
1–5 6–20 21–35 36–60 61–80
Frequency ( f )
24
9
21
15
42
Class width
(cw)
5
15
15
25
20
Frequency
density ( f/cw)
4.8
0.6
1.4
0.6
2.1
Frequency density
Number of
typing errors
2.0–3.5 3.5–4.5 4.5–5.5 5.5–7.0 7.0–9.0
Frequency
(f )
8
25
28
31
12
Class
width (cw)
1.5
1
1
1.5
2
Frequency
density
( f/cw)
5.33…
25
28
20.666…
6
Frequency density
x
iii E(X) = (0 × 0.4) + (1 × 0.24) + (2 × 0.144)
+ (3 × 0.216)
7
i The upper quartile is in the class ‘5.5–7.0’ cm.
30
25
20
15
10
5
0
5
2
3
4
5
6
7
8
9
Length (cm)
4
3
2
1
0
1
10 20 30 40 50 60 70 80
Number of errors
10 iThere must be a fixed number of trials.
Each trial must have exactly two discrete
possible outcomes, one of which is defined as
‘success’.
The trials must be independent with the
probability of success the same in each trial.
ii
µ ≈ (3 × 24) + (13 × 9) + (28 × 21) + (48 × 15) + (70.5 × 42) = 40.2 errors
111
1
9) + (28 × 21) + (48 × 15) + (70.5 × 42) = 40.2 errors
111
1
59
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SUMMARY REVIEW
ii X ~ B(18, 0.15)
Using the cumulative normal distribution
tables in reverse, we can say that P(Z < 1.406)
P(X  3) = 1 − P(X  2)
= 0.92. Therefore:
= 1 − [P(X = 2) + P(X = 1) + P(X = 0)]
c − 14.2
−1.406
=
 18
3.6
2
16  18 
17  18 
18 
= 1 −   × 0.15 × 0.85 +   × 0.15 × 0.8
85 +   × 0.85 
1
2
0






−5.0616 … = c − 14.2

 18

 18
 18
= 1 −   × 0.152 × 0.8516 +   × 0.15 × 0.8
8517 +   × 0.8518 
1
2
0








= 1 − 0.4796 …
= 0.520 (3 s. f.)
11 i P(2 paperback and 2 hardback) =
6 5 2 1
1
× × × =
8 7 6 5 28
c = 9.14 cm (3 s.f.)
ii Converting from x-values to z-values, we get:
16 − 14.2
z1 =
= 0.5 and z 2 = 15 − 14.2 = 0.2
3.6
3.6
P(15 < X < 16) = P(X < 16) − P(X < 15)
= P(Z < 0.5) − P(Z < 0.22…)
= 0.6915 − 0.5879
But there are 4C2 = 6 ways of selecting the two
hardback books within the selection of 4.
Therefore, P(X = 2) = 1 × 6 = 3
28
14
= 0.1036
So X ~ B(7, 0.1036)
P(X  2) = 1 − P(X  1)
= 1 − [P(X = 0) + P(X = 1)]
Alternative method: There are 8C4 = 70 different
6
2
possible selections. C2 × C2 = 15 of them
 7

 7
= 1 −   × 0.89647 +   × 0.1036 × 0.89646 
involve exactly 2 hardback books. So the
1
0
 
 

probability is 15 = 3
70
 7

14
 7
= 1 −   × 0.89647 +   × 0.1036 × 0.89646 
ii P(X = 0) = P(X = 1) = 0 as they are not
 1
 0

possible.
= 1 − 0.8413 …
P(X = 2) = 3 from part (i)
14
= 0.159 (3 s.f.)
6
5
4
3
360
3
P(X = 4) = × × × =
=
13 i If there are more women than men, then the
8 7 6 5 1680 14
number of women must be 4, 5 or 6.
Since probabilities sum to 1, we can say that
8C × 5C = 700
For 4 women and 2 men:
4
2
P(X = 3) = 8 .
8C × 5C = 280
14
For 5 women and 1 man:
5
1
For 6 women and 0 men:
So the probability distribution table is:
x
2
3
4
P(X = x)
3
14
8
14
3
14
iii Var(X) = E(X2) − E2(X)
=  22 × 3 + 32 × 8 + 4 2 × 3  − 32

14
14
14 
132
−9
14
= 0.429 (3 s.f.)
=
12 i
5C
0 = 28
Therefore, the total number of ways is:
700 + 280 + 28 = 1008
ii The committee could include one of the two
males (M1 or M2) or neither of them.
For M1 selected and M2 unavailable, we must
then select 2 men from the remaining 3, and
3 women from the remaining 8. So there are
3C × 8C = 168 arrangements.
2
3
There are another 168 arrangements for M2
selected and M1 unavailable.
If neither of the men are selected, we must
then select 3 men from the remaining 3, and
3 women from the remaining 8. So there are
3C × 8C = 56 arrangements.
3
3
8%
60
6×
8C
So the total number of ways is:
168 × 2 + 56 = 392
c
57767_P057_066.indd 60
14.2
7/5/18 4:09 PM
WORKED Solutions
15 iThere must only be two possible outcomes
(success and failure). Trials must be
independent. The probability of success must
be the same in each trial.
iii If we assume that both ends of the line are
occupied by women, then there are 5P2
ways of choosing those women. There are 4!
ways of arranging the middle 4 committee
members regardless of their gender.
ii P(banana on the 4th day) = 0.723 × 0.28
= 0.105 (3 s.f.)
P2
5


16 P( 25 < X < 30 ) = P  25 − 28.3 < Z < 30 − 28.3 

4.5
4.5 
= P(– 1.556 < Z < 0.801)
= P(Z < 0.801) – (1 – P(Z < 1.556))
= P(z < 0.801) + P(Z < 1.556) – 1
4!
= 0.7884 + 0.9401 – 1
So the number of ways = 5P2 × 4! = 480
= 0.729 (3 s.f.)
Alternatively, 5! × 4 = 480
14 iIf the coin shows heads, then we must use
the sum of the two dice, each of which has a
minimum score of 1, so X = 1 is impossible
with heads. With tails, X = 1 is possible if the die
shows a score of 1.
P(X = 1) = 1 × 1 = 1
2 4 8
ii To find P(X = 3), we need to consider (H, 1, 2),
(H, 2, 1) and (T, 3).
1 1 1
1 1 1
1 1
P(X = 3) = 2 × 4 × 4 + 2 × 4 × 4 + 2 × 4
(
)(
)(
)
17 iThe probability of arriving on time is 0.2 so
X~B(96, 0.2).
E(X) = 96 × 0.2 = 19.2 and
Var(X) = 96 × 0.2 × 0.8 = 15.36
So a normal approximation is X~N(19.2, 15.36)
For the normal approximation of P(X < 20),
the continuity correction means that we use
X = 19.5


P( X < 20 ) = P  Z < 19.5 − 19.2  =

15.36 
P(Z < 0.0765) = 0.531 (3 s.f.)
iiLet E be early, OT be on time, L be late and B be
eat a banana.
= 1 + 1 + 1
32 32 8
= 3
16
P(E) = 0.05
iiiTo find P(X = 2), we need to consider (H, 1, 1),
and (T, 2).
1 1
1 1 1
P(X = 2) = 2 × 4 × 4 + 2 × 4
1 1
=
+
32 8
5
=
32
To find P(X = 4), we need to consider (H, 1, 3),
(H, 2, 2), (H, 3, 1) and (T, 4).
∴ P(B′ ∩ E) = P(B′|E) P(E) = 0.3 × 0.05 = 0.015
P(B | E) = 0.7
(
)(
(
)
)
(
)
P(X = 4) = 1 × 1 × 1 × 3 + 1 × 1 = 7
2 4 4
2 4
32
Since probabilities sum to 1, P(X = 6) = 3
32
P(B|OT) = 0.4
⇒ P(B′ | E) = 0.3
⇒ P(B′|OT) = 0.6
∴ P(B′ ∩ OT) = P(B′|OT) P(OT) = 0.6 × 0.2 = 0.12
P(B|L) = 0
⇒ P(B′|L) = 1
∴ P(B′ ∩ L) = P(B′|L) P(L) = 1 × 0.75 = 0.75
P(B′) = P(B′ ∩ E) + P(B′ ∩ OT) + P(B′ ∩ L)
= 0.015 + 0.12 + 0.75 = 0.885
P(OT ∩ B′) = 0.2 × 0.6 = 0.12
P (OT |B ′ ) =
P(OT ∩ B ′) 0.12
=
= 0.136
P(B ′)
0.885
So the completed table is:
x
1
2
3
4
5
6
7
8
P(X = x)
1
8
5
32
3
16
7
32
1
8
3
32
1
16
1
32
iv P(Q ∩ R) = 0 since if the coin lands on tails,
then the 1 remaining die cannot score 7.
Therefore Q and R are mutually exclusive.
61
57767_P057_066.indd 61
7/5/18 4:10 PM
Summary REVIEW
Alternative method
0.7
B
0.3
0.05
E
0.75
L
0.2
OT
B’
 n 
0
n
 0  × 0.2 × 0.8 < 0.01


0.8n < 0.01
log 0.8n < log 0.01
0
B
1
B’
0.4
B
0.6
nlog 0.8 < log 0.01
n > log0.01 (change of sign since log0.8 < 0)
log0.8
n > 20.63 …
Therefore, the lowest possible value of n is 21.
20 X~Geo(0.35)
⇒ E ( X ) = 1 = 2.85… ⇒
0.35
B’
P(OT ∩ B ′)
0.2 × 0.6
Daniel should expect to ask 3 people.
P(OT | B ′) =
=
P(B ′)
(0.05 × 0.3) + 0.75 + (0.2 × 0.6)
21 iX ~ N(μ, σ 2)
P(OT ∩ B ′)
0.2 × 0.6
P(OT | B ′) =
=
P(B ′)
(0.05 × 0.3) + 0.75 + (0.2 × 0.6)
63 = 0.105 ⇒ P(X < 6) = 0.105
600
= 0.12 = 0.136 (3 sf
. .)
∴ From the tables, z = – 1.253
0.885
6−µ
2
∴
18 i E ( x − c ) = 1957.5 = 65.25
σ = −1.253 (1)
30
155 = 0.2583 ⇒ P(X > 12) = 0.2583
600
234
E( x − c ) =
= 7.8
∴ From the tables, z = 0.648
30
12 − µ
∴
(2)
Var(x) = Var(x – c) = 65.25 – 7.82 = 4.41 ⇒ σ = 2.1
σ = 0.648
(
)
ii ∑ ( x − c ) = ∑x − ∑c
234 = ∑x − 30c
∑x = 234 + 30c
234 + 30c = 86
30
234 + 30c = 2580
30c = 2346
c = 78.2
Alternatively: E(X) = E(X – c) + c
86 = 7.8 + c ⇒ c = 78.2
19 i T
here are 3 multiples of 5 in the given range and
15 possible numbers to choose from.
P(multiple of 5) = 3 = 0.2
15
(2) – (1)
6 = 1.901σ
⇒
σ = 3.156... = 3.16 (3 s.f.)
Substitute in (2)
μ = 12 – 0.648 × 3.156 = 9.95 (3 s.f.)
ii P(Z < – 1) = 1 – 0.8413 = 0.1587
P(Z > 1) = 1 – 0.8413 = 0.1587
P(Z < –1) or P(z > 1) is 2 × 0.1587 = 0.3174
0.3174 × 1000 = 317.4
⇒ 317 feathers
22 ai
There are 7! ways to arrange the 7 couples.
7! = 5040
Each couple has two possible
arrangements and there are 7 couples.
Therefore, the total number of
arrangements is 5040 × 27 = 645 210.
X~B(12, 0.2)
ii There are 7! arrangements of friends, 7!
arrangements of partners and the whole
ii P(3  X  5) = P(X = 3) + P(X = 4) + P(X = 5)
group can be arranged as (partners and
 12
 12
 12
5
P ( 3 X 5) =   × 0.23 × 0.89 +   × 0.24 × 0.88 +   × 0.2friends)
× 0.87 or (friends and partners). So the
 3
 4
 5
number of different arrangements is
 12
 12
 12
3
9
4
8
5
7
7! × 7! × 2 = 50 803 200.
3 X 5) =   × 0.2 × 0.8 +   × 0.2 × 0.8 +   × 0.2 × 0.8
 5
 3   4 
7! = 7 × 6 = 21
b i
ways
2!5!
2
P(3  X  5) = 0.236 … + 0.132 … + 0.053 … = 0.422
ii Adults all in team ⇒ only 1 way
iii X~B(n, 0.2)
Adults all not in team ⇒ 5! = 5 ways
4!1!
P(X = 0) < 0.01
So there are 6 ways in total.
62
57767_P057_066.indd 62
7/5/18 4:10 PM
WORKED Solutions
iii At least 2 girls means either 2 girls or 3 girls.
27 iThere are x yellow balls in box B (numerator).
3!
6!
6
×
5
The total number of balls in box B is 5 white
2 girls ⇒
×
= 3×
= 3 × 3 × 5 = 45
2!1! 2!4!
2
balls, x yellow balls and 1 additional white ball
3 girls ⇒ 1 of the remaining 6 people
= x + 6 balls (denominator).
with the 3 girls ⇒ 6 ways
So there are 51 ways in total.
ii
23 ∑ ( x − 36 ) = −60
∑x − 24 × 36 = −60
1
5
∑x = 804
∑x 2 = 27 011.76
24 i X~Geo(0.32)
P (Y1|Y2 ) =
P(X = 7) = 0.686 × 0.32 = 0.0316
ii P(X < 3) = P(X = 1) + P(X = 2)
= 0.32 + 0.68 × 0.32 = 0.5376
28 i
iii P(X > 3) = 0.683 = 0.314 (3 s.f.)


2

25 i Y ~N  µ, µ  P (Y < 0 ) = P  Z < 0 − µ  = P Z < − 2µ
µ 
µ
 4


2 


 µ2 
0 − µ
2µ

Y ~N  µ, P
Y
<
0
=
P
Z
<
(
)
µ  = P Z<− µ
 4 


2 
(
)
= P(Z < – 2) = 1 – P(z < 2) = 1 – 0.9722 = 0.0278
253
ii
P ( X > 2.1) =
= 0.03163
8000
5
x+6
White
x+1
x+6
Yellow
6 = 1 ⇒ x + 6 = 18 ⇒ x = 12
x+6 3
iv P(Y2) = P(W1 ∩ Y2) + P(Y1 ∩ Y2)
= 4 × 12 + 1 × 13 = 48 × 13 = 61
5 18 5 18 90 90 90
2
(
Yellow
iii
∑x − 72 × 804 + 24 × 36 = 227.76
2
x
x+6
Yellow
∑ ( x − 36 )2 = 227.76
∑x 2 − 72∑x + ∑362 = 227.76
White
White
4
5
∑x − ∑36 = −60
6
x+6
1 13 13
P(Y1 ∩ Y2) 5 × 18 90 13
=
=
=
61 61
P(Y2)
61
90
90
Flat screen
6
5
7
9
6
7
1
4
5
9
5
8
5
6
2
1
9
7
4 10
Key:
6|7|1
)
6
Conventional
4
7
means 0.76 m for flat screen
0.71 m for conventional
ii The median is 0.74 m
∴ From the tables, z = 1.858
IQR = UQ − LQ =  0.77 + 0.85  −  0.67 + 0.69  = 0.81 − 0.68 = 0.13 m

 

2
2
∴ 2.1 − 2.04 = 1.858 ⇒ σ = 0. 0323 (3 s.f.)
σ
IQR = UQ − LQ =  0.77 + 0.85  −  0.67 + 0.69  = 0.81 − 0.68 = 0.13 m



 2
2
26 i P(X < 73) = 0.15
∴ From the tables, z = –1.036
iiiUsing x to represent the diagonal lengths of flat
∴ 73 − 75 = −1.036 ⇒ σ = 1.93 (3 s.f.)
screen TVs: ∑x = 8.34 ⇒ µ = 8.34 = 0.927m
σ
9
ii P(X > 77) = 0.15 (by symmetry) ⇒ Y~B(8, 0.15)
7.7984
∑x 2 = 7.7984 ⇒ σ = 9 − 0.9272 = 0.0882m
P(Y < 3) = P(Y = 0) + P(Y = 1) + P(Y = 2)
 8
 8
 8iThere must only be two possible outcomes
29
P (Y < 3) =   × 0.150 × 0.858 +   × 0.151 × 0.857 +   × 0.152 × 0.856
 0
1
 2 (success and failure). Trials must be
independent. The probability of success must be
 8
 8
 8
the same in each trial.
P (Y < 3) =   × 0.150 × 0.858 +   × 0.151 × 0.857 +   × 0.152 × 0.856
1
 0
 2
ii P(buys on the nth call) = (1 – p)n–1p
P(Y < 3) = 0.272 … + 0.384 … + 0.237 … = 0.895
63
57767_P057_066.indd 63
7/5/18 4:10 PM
Summary REVIEW
⇒
ii If the number of doctors taking their final
examinations is an even number, then the
statement is always true, as the median is
p2 – p + 0.2016 = 0
positioned such that 50% of the data is below
1 ± 1 − 4(1)(0.2016) 1 ± 0.44
the median and 50% is above the median.
p=
=
2
2
If the number of doctors is odd, then the
p = 0.28 or p = 0.72
number of doctors above and below the
So the maximum probability is 0.72.
mean is equal, but not quite 50%, since the
data is discrete and the middle item of data
30 i X~B(20, 0.05)
is effectively discounted. However, as the
P(X > 1) = 1 – P(X  1) = 1 – P(X = 0) – P(X = 1)
number of doctors increases, each ‘half’ of the
 20
 20
0
20
1
19
P ( X > 1) = 1 −   × 0.05 × 0.95 −   × 0.05 × 0.95 data approaches 50%. For a large number of
0
1 
doctors, the data can be treated as continuous
20


 20
and so we would say that the statement is true.
> 1) = 1 −   × 0.050 × 0.9520 −   × 0.051 × 0.9519
1 
0
iii This scenario could be true in many situations.
P(X > 1) = 1 – 0.358 … – 0.377 … = 0.264
For example, suppose every doctor scored a
different mark, it would require two or more
ii Income = 450 × 10 = $4500
doctors just above 50% of the data to score
Cost = 24 × 20 = $480
the same mark and the statement is true.
If X > 1, there is a ‘profit’ of –$480
The mode is not an appropriate measure of
If X  1, there is a profit £4020
central tendency for this type of statement.
3 Typically, this would be a binomial distribution
So the expected profit
with X ~ B(275, 0.9). However, the large numbers
= (0.736 × 4020) + (0.264 × – 480) = $2832
make it impractical to use the binomial formula.
We know that:
Extension Questions
np = 275 × 0.9 = 247.5 and npq = 275 × 0.9 × 0.1
1 Consider the arrangement GGBBB. If the two girls
= 24.75
stay in the same position, there are 3! arrangements
Since np  5 and nq  5, we can approximate the
of the boys. We also know that the following
binomial distribution by a normal distribution,
arrangements of the girls will mean they are
X ~ N(247.5, 24.75).
together: BGGBB, BBGGB, BBBGG, GBBBG. The last

250 − 247.5 
P(X > 250) = P  Z >

arrangement works because they are standing in a

24.75 
circle and the two ends meet. In each case, there will
= P(Z > 0.502 518 …)
be 3! arrangements for the boys.
= 1 − P(Z < 0.502 518 …)
Therefore, the number of arrangements for the two
= 1 − 0.6924
girls standing together is (3! × 5).
iii When n = 2
0.2016 = (1 – p)p
0.2016 = p – p2
2
The total number of arrangements for the 5
children is 5!
Therefore, the probability of the girls standing
3! × 5 1
next to each other is
= .
5!
4
iThe spread of the data is the key factor if the
average referred to is the mean. This statement
is true if the distribution is symmetrical. It is also
true if the average distance from the mean for
top 50% of the data is equal in magnitude to the
average distance from the mean for the bottom
50% of the data.
4
= 0.3076
i T
he vertical scale begins at 51 000, which gives the
impression that more than twice as many people
voted for yes than no. To improve, raw data could
be used or the vertical axis could start at zero.
ii The diagram suggests that there is a
relationship between donations to charity and
mobile phone sales. The scales in each case
are different, so there may not be a correlation.
Also, correlation does not imply causation.
To improve, the data is probably best shown
on two separate diagrams with a scale.
64
57767_P057_066.indd 64
7/5/18 4:10 PM
WORKED Solutions
5
3C × 17C = 136 possibilities for choosing 3 of
3
2
the top 3
iii The average referred to here is the median,
which is not appropriate since it does not
take into consideration sugar values that
are significantly greater than the median.
To improve, base the calculation on the mean
for all sweets, also considering the probability
of each sweet being chosen.
1
P(1st person is selected) = n
P(2nd person is selected) = n − 1 × 1 = 1
n
n −1 n
P(3rd person is selected)
= n − 1 × n − 2 × 1 = 1
n
n −1 n − 2 n
(2040 + 136)
= 0.140 (3 s.f.)
15504
10 ip + 2pq + 3pq + 4q = 2.5 ⇒ p + 5pq + 4q = 2.5 (1)
p + pq + pq + q = 1
p + 2pq + q = 1
p + 2pq = 1 – q
p(1 + 2q) = 1– q
1−q
1 + 2q (2)
Substitute (2) in (1):
1−q
1−q
+ 5q
+ 4q = 2.5
1 + 2q
1 + 2q
p=
his pattern continues for all n people regardless
T
of height. Since every person has an equal
chance of being selected, we can say that the
probability of selection is independent of height.
6
1 – q + 5q(1 – q) + 4q(1 + 2q) = 2.5(1 + 2q)
1 – q + 5q – 5q2 + 4q + 8q2 = 2.5 + 5q
3q2 + 3q – 1.5 = 0
2q2 + 2q – 1 = 0
For the first digit, there are 8 possible options.
The question tells us that 5 is not allowed and we
also know that to be a 5-digit number the first digit
cannot be 0.
q=
−2 ± 4 − 4(2)(−1)
4
q = −2 ± 12 = −2 ± 2 3 = −1 ± 3
4
4
2
0q1
⇒
q = 3 −1
2
Therefore, the number of combinations is


8 × 9 × 9 × 9 × 9 = 52 488.
1 −  32− 1 
2− 3 −1


3
3 = 3 3 − 3 = 3 −1
ii p =
=
= −
6
2
7 a+b = 2 ⇒ a+b=4 ⇒ b=4–a


2 3
3
1
−
2
2
3
1
+
−
2
1 + 2
2 

( a − 2)2 + (b − 2)2 = ( a − 2)2 +1(−2− a3)2− =1 a 2 − 4a + 4 + 4 − 4a + a 2 = 2a 2 − 8a + 8
σ2 =
2 − 32− 1
2
2  2 
2
=
p=
= 3− 3 = 3 3 − 3 = 3 −1
2
2
2
2
2
2
6
2
2 3
1
(b − 2) = ( a − 2) + ( 2 − a ) = a − 4a + 4 + 4 − 4a + a = 21a +−28 a +3 8− 1  2 + 2 3 − 

2
2
2
2
2

 3 − 1  3 − 1 3 − 2 3 + 1 4 − 2 3 2 − 3
=
=
=
iii pq = 
2
2
2
4
2
4
 2   2 
σ = a – 4a + 4 = (a – 2)
For the remaining 4-digits, the only disallowed
number is 5, so there are 9 choices in each case.
(
(
(
(
σ = |a – 2|
8
9
)
)
)
)


 pq =  3 − 1   3 − 1  = 3 − 2 3 + 1 = 4 − 2 3 = 2 − 3
2
2
4
4
2



i 1 × 6 × 5 × 4 × 3 = 360
7 7 7 7 2401
2− 3
2− 3
3 −1
E X 2 = 3 −1 + 4
+9
+ 16
ii 1 − 360 = 2041
2
2
2
2
2401 2401
2 or more of the top 3 students ⇒ 2 students or 3
2 − 3 2− 3
3 −1
E X 2 = 3 −1 + 4
+9
+ 16
students in the group of 5
2
2
2
2
For 2 students:
E X 2 = 3 − 1 + 8 − 4 3 + 18 − 9 3 + 16 3 − 16 = 9 + 4 3
244800
2
2
3 × 2 × 17 × 16 × 15
× 5C 2 =
20 × 19 × 18 × 17 × 16
1860480
−
+
−
+
−
+ 16 3 − 16 = 9 + 4 3
3
1
8
4
3
18
9
3
2
E X =
2
2
For 3 students:
 9+4 3
16320
3 × 2 × 1 × 17 × 16
5
– 2.52 = 1.71 (3 s.f.)
Var ( X ) = 
× C3 =
2 
20 × 19 × 18 × 17 × 16
1860 480

( )
( )
So the required probability is:
24480
26112
1632
+
=
= 0.140 (3 s.f.)
1860 480 1860 480 1860 480
Alternatively:
20C possibilities = 15 504
5
3C × 17C = 2040 possibilities for choosing 2 of
2
3
(
( )
) ( )
(
(
) (
)
(
)
)
( )
11 i Let n be the number of red balls. Then (n + 2)
is the number of green balls and (2n + 2) is the
total number of balls.
P( different ) =  n × n + 2  +  n + 2 × n 
 2n + 2 2n + 2   2n + 2 2n
n + 2
the top 3
65
57767_P057_066.indd 65
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Summary REVIEW
∴ 2n(n + 2)
= 195
(2n + 2)2 392
784n(n + 2) = 195(2n + 2)2
784n2 + 1568n = 195(4n2 + 8n + 4)
784n2 + 1568n = 780n2 + 1560n + 780
4n2 + 8n – 780 = 0
n2 + 2n – 195 = 0
(n + 15)(n – 13) = 0
n = –15
or
n = 13
n>0
⇒
n = 13
Therefore, there are 13 red balls and 15 green
balls.
494 325
ii P(RGRGR) = 13 × 15 × 13 × 15 × 13 =
28 28 28 28 28 17 21
10 368
570 375
15
13
15
13
15
×
×
×
×
=
and P(GRGRG ) =
28 28 28 28 28 17 21
10 368
570 375
15
13
15
13
15
P(GRGRG ) =
×
×
×
×
=
10 368
28 28 28 28 28 17 21
494 325
570 375
1 064 700
38025
P ( alternating ) =
+
=
=
17 210 368 17 210 368 17 210 368 614656
494 325
570 375
1 064 700
38025
rnating ) =
+
=
=
17 210 368 17 210 368 17 210 368 614656
12 P( X = x ) =
n− x
n!
p x (1 − p )
n
x
x
!
!
−
( )
P ( X = x − 1) =
P( X = x )
=
P ( X = x − 1)
(
n −(x −1)
n!
p x −1 (1 − p )
n − ( x − 1) !(x − 1)!
)
n− x
n!
px 1 − p
(n − x )! x ! ( )
n −(x −1)
n!
p x −1 (1 − p )
(n − ( x − 1))!(x − 1)!
n ! ( n − x + 1)! ( x − 1)! p (1 − p )
P( X = x )
=
n − x +1
P ( X = x − 1)
n ! ( n − x )! x ! p x −1 (1 − p )
x
n− x
P( X = x )
p(n − x + 1)
=
x(1 − p)
P ( X = x − 1)
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