+ 1.5 x IQR
+ 1.5 x 30.5
b upper boundary = Q3
SOLUTIONS TO TOPIC 5
(STATISTICSAND PROBABILITY)
= 137.5
= 183.25
lower boundary = Q1 - 1.5 x IQR
NO CALCULATORS
= 107 - 1.5 x 30.5
= 61.25
a There are 72 cars in total.
4
+ 11 + x + 18 + 12 + 6 + 4 + 2 =
.'. x + 57 =
There are no data values which are either greater than
upper boundary or less than the lower boundary, so
are no outliers.
72
72
.'. x = 15
b The modal class is 60 ~ v
c
v
30 ~ v
40 ~ v
50 ~ v
60 ~ v
70 ~ v
80 ~ v
90 ~ v
100 ~ v
<
<
<
<
c
70 km h-1.
Frequency
Cumul. Frequency
o
o
4
4
15
30
48
60
66
70
72
< 30
< 40
< 50
< 60
<
<
11
15
18
12
6
4
2
70
80
90
100
110
•
It
I
90
I
I
I
I
I
I
I
•
100 110 120 130 140 150 160
a From the graph, 100 units were sold for < $200 000.
b
When
N = ~ of 800 = 400,
the value = 300 thousand
the median = $300 000
c IQR = Q3 - Q1
= $375 000 - $250000
= $125000
cumulative frequency
70
60 /54
---.----------------------------------50
40
30
20
I
80
d From the graph, 500 units were sold for
300 units were sold for> $330000.
.. -----------
P(selling price>
'·29..·································
'·is..···························
5
$330000)
=
~gg=
<
$330 000..
~
a
10
10
20
30
median =
40
60
q,~::::;63
ii Q1::::; 52,
.'.
56
km h
70
80
90 100 110
"!
QiJ::::; 75
IQR = Q3 - Q1
::::;75 - 52::::; 23 kmh-1
iii 40% of 72 ::::;29,
The line of best fit passes through (120, 22) and (80,
so the 40th percentile se 59 km h -1.
The line has gradient
2
total
mean = -= 11
7
a
7
+ 9 + 9 + x + 13 + 13 + 16
7
.'.
x
= -0.15
Its equation is
=11
+ 67
= 77
.'. x
=
12 = 77
.'.
10
b
y = -0.15(100)
ii y = -0.15(200)
+t
C
8
96 = 77
= 19
y- 28
x _ 80 = -0.15
Y - 28 = -0.15x
b The total of the scores is 67 + 10 = 77.
If Kai scores t goals in the next game,
then
-
28 - 22
m = --80 - 120
+t
.'. t
y
=
+ 40
+ 40
-0.15x
+ 12
+ 40
= 25
= 10
The first calculation is likely to be more reliable, as it .
an interpolated value. 200 grams is outside the domain,
however so the second calculation is an extrapolation anc.
less reliable.
So, she must score 19 goals next time.
,.-A-...
3
~
,.-A-...
~
6
a
22 cm
b
range = 22 - 4
85 96 98 100 105 106 108 108 112 112 118 120 123
. 125 126 128 133 135 140 144 144 148 148 156
'-v-"
Q3
a
Q1 = 107
ii median = Q2 = 121.5
iii Q3 = 137.5
iv IQR = 30.5
v range = 156 - 85 = 71 .
Mathematics
SL - Exam Preparation & Practice Guide (3rd edition)
ii 4 cm
= 18 cm
iii 19 cm
iv 17 cm
ii IQR = 19 - 11
= 8 cm
C
75% of seedlings were taller than 11 cm.
d
No, the distribution is negatively skewed (skewed to the
left).
a
7
Die 2
6
5
4
@
1
b
0
P(sum more than 8)
5
4
@
2
1
0
1
= P(sum
4
@
2
1
0
1
2
10
3
2
1
0
1
2
@
2
1
0
1
2
@
4
1
0
1
2
@
4
5
1
b
2
2
4
3
5
6
36
5
18
ii
11 + 8
8 There are
*
Die 1
a
.2
P(T')
b
11 + 2
25
19
13
25
25
P(plays at least one sport)
23
13
a
8
19
P(T n E')
b
a f(x)?:
J:
0 on a given interval
a:::;; x :::;;b,
~
3
ii P(T U E') = 1 - P(T'
= l.
:::;;X2) = r
J
X2
f(x) dx
X1
n E)
=1-~xt
=1-~=~
a The random variable is discrete.
i+i+k+fi=l
b
= ~x
1
f(x) dx
b P(X1:::;; X
o
/5
=
8
11+8
25
and
6)
I H)
P(T
d
11+8+4
25
9
=
{1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1}
students.
11 +8
25
(
{marked x }
iii P(sum
P(H)
sum :::;;'7)
1
i'B = i.
+ 4 + 2 = 25
P(more than 3, but less than 8)
= P(4:::;;
There are 6 outcomes where the difference is 3.
As all outcomes are equally possible, the probability of the
difference being 3 is
?: 9)
14
.'. k+f2=l
. .'. k
= fi
"
( E(X)=-2(i)+0(i)+3(fi)+5(fi)=1
,,
b=71-a
c=44-a
a a+b=71
a+c.=44
a + d = 21
d
= 21-
a
a+ b+ c+ d =
a) + (44 - a) + (21- a) =
.'. 136 - 2a =
a=
So, b = 53, c = 26, d =
Now
a + (71-
P(at least one of each sex)
100
= ~= ~
100
100
15
a The scatter diagram suggests there is a very strong positive
relationship between age and arrnual income.
18
b The mean point is (27, 20).
3
The line of best fit passes through (27, 20) and (39, 30).
The line has gradient
m
=
30 - 20
39 _ 27
_
10
-12
3
UL.-
5
"6
---'
Its equation is
b
p(RnE')
ii P(R nE)
iii P(R
U
B)
= 11080=
=
=
0.18
6
5
4
~
3
!
2 ..........•..
_--J _-)j(......
i
.
··x
There are 36
possible outcomes
each shown by
a dot point.
x = 30,
When
~x -
¥
y = ~(30) - ~ = 22.5
So, the annual income is approximately
ii When
x = 60,
d
• .
5 6
••
$22500.
y = ~(60) - ~ = 47.5
So, the annual income is approximately
11--+-+-:~~*"i-*-1
.-+--'---'---'---'---'---'--....... Di e 1
1 234
=
.: y=~x-~
(
h~h~~:
=t::::;:~,d
__ sum > 9
5
6
-
x - 27
y - 20
= 0.53
18\"03l26
= 0.97
15~0
Die 2
a
y - 20
--
$47500.
The age of 30 is within the given data range but 60 is
outside the range. Predicting the income at 30 years is
an interpolation and is more reliable than the extrapolation
required for 60.
Mathematics
SL - Exam Preparation & Pradice Guide (3rd edition)
16
Tom
Jerry
0.7............H <::::: 0.60.4 __
<
0.3 .........•.
M <::::: 0.60.4 __
a
o + 4 + 6 + 3a + 8
2:.f
a x=
H
M,f
3a + 18
a+ 15
H ,f
M
3a
b
=
1.65
+ 18 = 1.65a + 24.75
=
6.75
a=--
6.75
1.35
1.35a
+ (0.3)(0.6)
= (0.7)(0.4)
= 1.65
6+4+3+a+2
P(only one of them is successful)
=
17
- 2:. fx
2
a=5
0.46
b The standard deviation ~ 1.39
{technology}
P (at least one is successful)
=
=
=
1 - P(both miss)
=
0.88
3 Using technology
a
1 - (0.3)(0.4)
J.l ~
1 - 0.12
4
a
90
3.56,
b
a ~ 0.512
J.l ~
67.9,
a ~ 12.2
+ 100 + 93 + 96 + p + 107 + 98 + 98 + 92 = 97
9
.
774+p=873
a
p
b
.§..-11
DB
5
LB
LB~!r-
.§..--
99
The ordered data is:
90, 92, 93, 96, 98, 98, 99, 100, 107
.'. the median is 98.
a The sum of the frequencies is 30, so 30 drives were chosen.
b
ft- w
=
8 fr equency
DB
6
LB
4
w
2
-
w~~1
rrb
P(2 whites)
ii
i6
= 12 x-&.=
P( different colours)
=
=
1-
[(f2)h~iJ + (~)(fI)
=
1-
12;:1l
1 - P(both the same colour)
a
= ~
a+ b+c+d=
a
Since a+ b
=
17,
d
Since b + c = 12,
BW'
3
+ b+ c =
26
230::::;d
235::::; d
+c=
26
240::::;d
245::::; d
17
b+9
=
250::::; d
255::::; d
260::::;d
12
b=3
<
<
<
<
<
<
<
<
Cumul. Frequency
1
235
240
3
1
4
5
7
9
16
3
6
4
19
25
29
1
30
245
250
255
260
265
a+b=17
9
a
U
+3 =
17
=
14
a
e ~ ~ 63.3%
The golfer hit a distance of more than 235 metres on 26
out of the 30 drives sampled.
a 3 students have both brown hair and blue eyes.
= fa =
= 12 = 0.25
P(BI but not Br)
ii P(Br I Bl)
Frequency
230
Distance (m)
d=4
4
b
h--,
225 230 235 240 245 250 255 260 265
distance (m)
225::::; d
c=9
14
v
Since the mean, median, and mode are about 'equal, the
data appears to be symmetric.
30
b + c = 12
d
U
--,
a+b=17
c
b
-
c An estimate of the mean is 245.
An estimate of the median is 243.
The modal class is 240::::; d < 245.
+ (12)(-&')]
B(ID'
18
o
-
So, if the golfer hit 100 drives, we would expect
100 x ~ ~ 87 drives to travel more than 235 metres.
0.3
6
CALCULATORS
a Using technology,
x=
7.52 marks
b The mode = 8 marks
c There are 25 data values, so the median is the
25i1
.'.
Mathematics
=
13th data value.
46
the median is 8 marks.
SL - Exam Preparation
& Practice
Guide
(3rd edition)
El
66
70
x
a P(X
>
peA) =:r;-
Let
60) ::::;0.309
0.63 = x
Over a 100 year period we would expect more than 60 mm
of rain to occur 31 times.
b P(50 < X < 60) ::::;0.669
Over a 100 year period we would expect between 50 mm
and 60 mm of rain to occur 67 times.
7 The tree diagram uses:
F for female
L for left-handed
M for male
R for right-handed
0.13
+ 0.36
- 0.36x
0.27 = 0.64x
.'. x = peA) ::::;0.422
=
12 P(AUE)
peA
U
.'. H
peA n E)
L
= 1- f2 = H
+ P(B) - peA n E)
= 0.46 + ~- peA n E)
= 0.46 + ~- H
1 - P((AUE)')
E) = peA)
peA nE) ::::;
0.258
3VF~R
13
2nd draw
a
~M~L
1st draw
R
0.76
<15
<:
+ (0.62)(0.24) ::::;0.198
n L) = (0.38)(0.13) ::::;0.249
b P(F I L) = P(F
P(L)
P
R
P
7/
a P(L) = (0.38)(0.13)
2."-
s ......•.
10~R<:
7..-P
..:l..
0.1982
9 ......•.
R
8
a r::::;0.898
There is a strong correlation between the age of contestants
and the time taken to complete the task.
b
peat least one red ticket)
= 1 - P(no red tickets)
= 1- P(PP)
=l-fox~
0.930x - 1.56
by::::;
ii We expect that an increase of one year in age will
add 0.93 minutes to the time to complete the task.
= 1- ~
9
ay::::;
1.09x - 0.781
b
0.883
r::::;
48
{using technology}
96
ii
c There is a strong positive correlation between the times
=~+~
cd Subject D is an o~t!ier.
e y::::; 1.094x - 0.7S<i3
42
96
y = 48 gives
48::::;1.094x - 0.7813
iii
=~+~
63
96
a r::::;0.995
14
b There appears to be a very strong, positive correlation
between monthly rainfall and crop yield.
c The least squares regression line is
y::::; 0.606x
d
+ 9.4
If x =0,
then
y::::; 0.606(0)
+ 9.4.
20
y::::; 0.606(12)
+ 9.4
o
11
So, the expected yield is approximately
data range; we expect it to be unreliable, especially as it
predicts such a large yield when there is no rain at all.
In contrast, the estimate in ii is an interpolation, and we
would expect the estimate to be reliable.
= peA) + P(E) - peA n E)
peA U E) = peA)
so peA
+ P(E)
n E)
12
/:
time (minutes)
~ ~
13
14
15
16
17
16.7 tonnes.
e The estimate in i is an extrapolation as it is outside the
Hence,
------
1~
9.4 tonnes.
::::;16.7
But A and E are independent,
,(
I:
15
So, the expected yield is approximately
11 peA U E)
7
16
Finishing times
35 .-----:-:---;:------------cumulative frequency
30
~-25 ~--------------/7~~---------
::::;9.4
ii If x = 12, then
or RP)
=fox~+fox~
x::::; 44.6
So, we expect that an adrenaline injection would reduce the
mouse's time to 44.6 s, a reduction of about 3.4 seconds.
10
7
15
P(purple ticket second)
= P(PP
48.7813::::; 1.094x
.'.
P(one ticket of each colour)
= P(PR or RP)
=fox~+fox~
taken to complete the obstacle course of mice without
adrenalin~ and the time'S of mice with adrenaline.
Substituting in
8
15
a median re 15th score se 13.8 (or 13.9) minutes
b
7~th score se 13.3 min.
So, any time less than 13.3 minutes.
c 2 finished within 12 minutes and 25 finished within 15 min
.'. 23 finished between 12 and 15 minutes.
d
= peA) P(E).
Ql::::;
5 runners out of 30 finished within 13 minutes.
Pea runner finished in less than 13 min) =
fa-
= ~
- peA) P(E).
El
Mathematics
SL - Exam Preparation
& Pradice
Guide (3rd edition)
15
J.L= 310 mL,
(J"= 5 mL
ii The relationship has changed from moderate to very
strong.
a P(300 < X < 310) ~ 0.477
.'.
47.7% lie between 300 mL and 310 mL.
b P(X?
••
+ 32.0
a y R:::2.43x
20
304) ~ 0.885
b If Y
88.5% are at least 304 mL.
=
70 R:::2.43x
38 R::: 2.43x
70,
e P(X < 300) R:::0.0228
+ 32
.'. x ~ 15.6
So, we estimate that Tony revised for 15.6 hours.
16
For English, P(X > 80) R:::0.266
.'. 26.6% scored higher than Ashleigh
a
e The y-intercept (32%) is the estimate of the result for a
student who did not do any revision.
ii For History, P(X > 72) R:::0.212
.'. 21.2% scored higher than Ashleigh
b z-score for English
80;
The gradient of the line indicates that the result will
increase by 2.43% for each additional hour studied.
75 = 0.625
21
72 - 60 = 0.800
15
Ashleigh achieved a higher standard in History.
z-score for History
.'.
17
Let X be the number he correctly guesses, so
X ~ B(10, 0.25)
a
X ~ N(J.L, (J"2)
Now
>
90)
=
0.12
P(X ~ 90)
=
0.88
=
0.88
P(X
P ( X ; J.L~ 90;
J.L)
P(Robert fails)
= P(X ~ 2)
R:::0.526
e
=
J.L)
0.88
<
P ( X ; J.L < 60;
<
60
P(X = 3)
R:::0.250
5)
R:::0.0781
(J"
P (Z
=
1- P(X ~ 4)
R:::1 - 0.9219
22
90 - J.LR:::1.1750(J" .... (1)
P(X
P( answers exactly 8 correctly)
P(achieves a C or better)
= P(X?
90 - J.LR:::1.1750
Also,
b
=
(z ~ 90;
P
If Robert gets the 5 known correct then there are 10 answers
he guesses.
=
60)
J.L)
;
.
J.L)
J.L=O,
(J"=1,
and X~N(0,12).
a P(X ~ 1) R:::0.841
0.20
b P( -0.5
=
0.20
e P(X
=
0.20
~ X ~ 0.5) R:::0.383
>
2) R:::0.0228
we can expect that 0.0228 x 850 R::: 19 speeds
measured will exceed the actual speed by more than
2 kmh-1.
60 - J.L';'R:::
-0.8416
U J.L=40,
(J"
60 - J.LR::: -0.8416(J"
.... (2)
a
Solving (1) and (2):
+ J.LR:::
Adding,
30
P(X
>
45) R:::0.159
<
X
<
50) R:::0.819
b We need to find k such that
P(X>k)=O.l
0.8416(J"
R:::2.0166(J"
P(X
30
(J"R:::-R:::14.88
2.0166
and
and X~N(40,52).
ii P(35
90 - J.LR::: 1.1750(J"
-60
(J"=5,
~ k) = 0.9
.'. k R:::46.4
40
90 - J.LR:::1.1750 x 14.88
So, the minimum
46.4 cm.
J.LR:::72.5
Thus : J.LR:::72.5 and (J"~ 14.9
k
• x
length of the longest 10% of fish is
24 J.L= 500, (J"= 2.5
18 p=t=0.8
a P(X < 495) R:::0.0228
Let X be the number of people sampled who oppose the traffic
lights.
X ~ B(20, 0.8)
a P(X
=
25
16) R:::0.218
b P(X?
=
16)
1 - P(X
R:::1-
~ 15)
0.370
R:::0.630
19
. air
b
ii y R:::-0.555x
R:::-0.647
Mathematics
+ 71.3
The outlier is (50, 12).
With the outlier removed:
i r R:::-0.986
ii y R:::-0.637x
e
&
Pradice Guide (3rd edition)
J.L= 38.4,
a
(J"= 4.6
i P(X
>
+ 80.5
43.5) ~ 0.134
ii P(X
~ 36.4) R:::0.332
iii P(30 ~ X ~ 40) R:::0.602
b We need to find k such that
P(X
> k)
P(X ~ k)
.'.
The gradient of the line is steeper (has become more
negative).
SL - Exam Preparation
b 0.0228 x 10000 R:::228 bottles will require extra sauce.
= 0.9
=
0.1
k R:::32.5
k
38.4
So, 90% will reduce their oxygen consumption
than 32.5 mL.
by more
a
26
P(X
<
56) = 0.8
p( Z
<
56;
J.L)
iii Of the 779 patients who were 50 or older, 469 were
male.
means that
=
0.8
32
Let X be the number of allergic reactions.
:.
X
B(5000, 0.1)
rv
<
P(X
56 - J.L~ 0.8416(4)
then from a,
=
470)
56 - J.L~ 3.366
:.
27
J.L= 1020,
a P(X
a
<
=
J.L~ 52.6
P(X
~ 469)
~ 0.0743
33
15
The length X cm of a chopstick is normally distributed with
mean J.Lcm and standard deviation 0.5 cm.
<
p(X-J.L
0.5
p(z<
2.28% of the containers overflow.
so n = 10, P = 0.88.
B(10, 0.88)
rv
a P(X
b
24-J.L)
0.5
=0.01
J.L~ 25.2 cm
34
= 8) ~ 0.233
=
? 8)
1 - P(X
If X cm is the length of a steel rod then X is normally
distributed with mean 13.8 cm.
Now P (X < 13.2) = 0.015
P ( X -a13.8
~ 7)
<
~1-0.109
rv
B(6, 0.05)
so n
=
~ 2)
So, the manufacturer will
have to pay double money
back on 0.223% of boxes.
~ 1 - 0.99777
~ 0.00223
P(x)
= P(X
Defective
Not defective
Total
37
581
618
Corn
Pineapple
24
617
641
Total
61
1198
1259
= x) = f4(x
3
=
b E(X)
~
0.0485
1~~9 ~
P(corn
I defective)
~
a 0.05
0.607
iii 37 out of 618 corn tins are defective.
:.
P(defective
I corn)
~
< 50
? 50
Total
Male
136
469
Female
155
310
605
465
Total
291
779
1070
31
7
631 8 ~
=
Xi)
+ 2(~) + 3(f4)
50
24
=212
36
*~
P(X
= 1(-k)
ii 37 of the 61 defective tins are corn.
.'.
L: Xi
i=1
61 out of the 1259 tins are defective.
P(defective)
x E {I, 2, 3}
= f4(9) = f4
a 1259 tins
.'.
+ 6),
P(2) = f4(8) = ~
_
b
~ -2.17009
-k
a P(l) = f4(7) =
P(3)
30
0.015
a ~ 0.276 cm
35
P(more than 2)
=
-0.6
a
6, PT 0.05.
The manufacturer must pay dquble money back if more than
two items are defective.
~ 1 - P(X
= 0.015
13.2 ~ 13.8)
p( Z < -~.6)
~ 0.891
29 X
0.01
24 - J.L ~ -2.326
0.5
24 - J.L~ -1.163
A student passes if they pass 8 or more of the tests.
P(X
=
=0.Q1
So, in a sample of 1500, we expect 0.6563 x 1500 ~ 984
containers will hold between 1 litre and 1.03 litres.
X
24)
24-J.L)
0.5
c P(1000 ~ X ~ 1030) ~ 0.6563
28
<
P (X
Now
1000) ~ 0.0912
b Containers that overflow have more than 1050 mL added
to them.
Now P(X > 1050) ~ 0.0228
:.
0.602
~!~
~0.667
I female) =
.'. P(50 or older
So, a score of 56 is 0.842 standard deviations from the
mean.
=4
= ~~~~
or older)
iv Of the 465 female patients, 310 were 50 or older.
56 - J.L ~ 0.8416
a
56 - J.L~ 0.8416a
b If a
I 50
:. P(male
+ k + 0.5 + 0.3 = 1
k + 0.85 = 1
k = 0.15
b E(X)
=
0.0599
=
=
L: xip,
0.15
= 2.05
a 1070 patients
37
0.3
+ 1(0.15) + 2(0.5) + 3(0.3)
+ 1 + 0.9
0(0.05)
+ 0.2 + m + n = 1
.'. m+n
= 0.5
.'. n = 0.5 - m
b
605 of the 1070 patients were male.
:. P(male)
=
6
1 0°;0 ~
ii 291 of the 1070 patients were younger than 50.
:. P(younger than 50)
=
i/o ~
2
1
E(X)
Now
0.565
0.272
L: x.p,
=
0(0.3)
.'.
1.55
.'.
1.35 = 2m
.'.
El
=
+ 1(0.2) + 2m + 3n
+ 3(0.5 - m)
m = 0.15
Mathematics
SL - Exam Preparation
& Pradice
Guide
(3rd edition)
38
a
r;:::::-0.832
b
As the number of tomatoes in a bag increases, the median
weight of tomatoes in the bag decreases.
=
-
1'(x)
Now
=
There is a moderate negative linear correlation between the
variables.
c
d
If x
=
13, then
y;::::: -8.21(13)
+ 224
ii If x
=
20, then
y;::::: -8.21(20)
Iim
+
224
So, the median weight is approximately
lim (7 - 2x - h)
h->O
=
7- 2x
60.2 g.
.'.
the point of contact is (1, 6).
Now
The estimate in i is therefore more
1'(1)
7 - 2(1)
=
y-6=5
x-I
= 0.613
J1.) = 0.613
P ( Z ~ 15 ~ 13)
which is
y - 6 = 5x - 5
or
= 0.613
2
+ h)
dy = lirn f(x
dx
h->O
= lim [-2(x
2
= lim -2(x
P (X ~ 24) = 0.035
h->O
=
0.035
= lim
24 - J1.;:::::
-1.812
a
24. - J1.;:::::
-1.8120-
= lim
.... (1)
p(z<
33-;J1.)
+ 2xh + h2)
h
0
+ 2X2
0
4xh - 2h2 ~
+ h)
-2j{(2x
0
+ h)
{as h
#
O}
dy
-=-4(-1)=4
dx
the tangent has gradient 4.
b When
=0.738
.'.
x=-l,
c When x=-l,
.'.
.... (2)
y=-2(-1)2+3=1
the point of contact is (-1, 1).
Since the gradient of the tangent is 4, the gradient of the
(1) - (2) gives
(24 - J1.)- (33 - J1.);:::::
-1.812a
.'. -9;::::: -2.449a
.'.
normal is -
- 0.6372a
i.
So, the equation of the normal is
y -1
1
x-(-l)
-"4
a;:::::3.67
+ 1.812(3.67)
;:::::30.7
So, the mean J1.;:::::
30.7 and the standard deviation
4y - 4 = - (x
a se 3.67.
.'. 4y
=
-x
.'. y=-4-
+ 1)
+3
-x+3
d The normal meets the curve where
SOLUTIONS TO TOPIC 6 (CALCULUS)
-x+3
-4HO CALCULATORS
f(x)=7x-x2
x - 9
-8X2
2
+3
+ 12
0
(8x-9)(x+1)=0
.'.
f(x + h) = 7(x + h) - (x + h)2
= 7x + 7h - (x2 + 2xh + h2)
= 7x
-
=-2x
+3=
=
-x
8x2
a
+ 3]
= -4x
= 0:738
33 - J1.;:::::
0.6372
a
33 - J1.;:::::
0.6372a
J1.;:::::
24
[-2x
h
h->O
"'.
2
+ h? + 3]h
A:? -
h->O
33) = 0.262
P (X < 33)
+1
- f(x)
h->O
= lim -2(2x
P(X?
Also,
5x
h
h->O
P ( Z ~ 24 -; J1.)
=
Y
y=f(x)=-2x2+3
a Let
2
- ;:::::0.287
a
a;:::::6.97
40
5
=
the tangent has equation
P(X ~ 15)
P ( X ; J1.~ 15;
{as h#O}
IX
b f(1)=7(1)-12=6
e The estimate in i is an interpolation whereas the estimate
39
h
lim X(7-·2x-h)
=
+ 7h
When
- X2 - 2xh - h2
x =~'
y =
x
=
-1 or ~
_2(~)2 +3
=-2(*)+3
15
32
So, the normal meets the curve again at (~'
Mathematics
+ X2
h
h->O
118 g.
- X2 - 2xh - h2 - 7x
7h - 2xh - h2
h->O
=
;:::::60.2
in ii is an extrapolation.
reliable.
+ 7h
7x
lim
;:::::118
So, the median weight is approximately
- f(x)
h
h->O
=
+ 224
y;::::: -8.21x
+ h)
f(x
Iim
h----tO
SL - Exam Preparation
& Pradice
Guide (3rd edition)
El
H).