S1 Mixed Starters
Height (mm)
Frequency
11-
2
16-
8
21-
15
26-
23
31-35
7
a) Calculate an estimate of the mean
b) Calculate the standard deviation of the data
Height
(mm)
f
X
fx
x²
fx²
11-
2
13
26
169
338
16-
8
18
144
324
2592
21-
15
23
345
529
7935
26-
23
28
644
784
18032
31-35
7
33
231
1089
7623
Σf = 55
Σfx² = 76520
Σfx = 1390
a) x = Σfx = 1390 = 25.3
Σf
55
b) Variance = Σfx² - Σfx
Σf
Σf
² = 76520 - 1390 ² = 752.56
55
Standard deviation = √752.56 = 27.43
55
Venn diagrams and conditional probability
P(C) = 0.4
P(D) = 0.7
P(C D) = 0.2
Calculate
a) P(D  C)
P(C`D`)
P (C`D)
P(CD) = P(CD) x P(D)
P(CD) = 0.2 x 0.7 = 0.14
C
D
0.26
0.14
S
0.56
0.04
Calculate
a) P(D  C) = 0.14 ÷ 0.4 = 0.35
P(C` D`) = 0.04
P (C`D) = 0.56
Weight (kg)
Frequency
10 - 14
20
15 - 19
14
20 - 24
4
25 - 29
1
30 - 34
1
a) Perform the calculations that you would need to be
able to draw a histogram (Do not draw the histogram)
b) Work out Q1, Q2 and Q3 and comment on the
skewness of the data set
c) Calculate D3 and P67
Weight (kg)
Frequency
10 - 14
20
15 - 19
14
20 - 24
4
25 - 29
1
30 - 34
1
Q1 = 12
Q2 = 14.5
Q3-Q2 = 3.6
Weight (kg)
Frequency
density
9.5 – 14.5
20÷5=4
14.5 – 19.5
14 ÷5=2.8
34
19.5 – 24.5
4 ÷5=0.8
38
24.5 – 29.5
1 ÷5=0.2
39
29.5 – 34.5
1 ÷5=0.2
40
Q3 = 18.1
Q2-Q1 = 2.5
Q3-Q2 > Q2-Q1 therefore there is positive skew
D3 = 12.5
P67 = 17.9
CF
20
The probability that is snows on a particular day in winter is
0.24.
If it snows the probability that I go to work is 0.18. If it does
not snow the probability that I go to work is 0.85.
a) Calculate the probability that it snows but I do not go to
work
b) Calculate the probability that I go to work
The probability that is snows on a particular day in winter is 0.24.
If it snows the probability that I go to work is 0.18. If it does not snow the
probability that I go to work is 0.85.
a)
Calculate the probability that it snows but I do not go to work
b)
Calculate the probability that I go to work
0.18
0.24
0.76
W
S
0.82
0.85
W`
W
0.15
W`
S`
a) 0.24 x 0.82 = 0.1968
a) (0.24 x 0.18) + (0.76 x 0.85) = 0.6892
Adam and Beth decide to play a game. The probability that
Adam wins the game is 0.2 and the probability that Beth
wins is 0.35. They decide to play the game 3 times.
Given that the results of the game are independents, find
a) P(wins all 3 games)
b) P(all 3 games end in a draw)
c) P(Adam wins 1 and Beth wins 2 games)
d) P(Adam and Beth win one game each)
Adam and Beth decide to play a game. The probability that Adam wins the game is 0.2 and the probability that Beth wins
is 0.35. They decide to play the game 3 times.
Given that the results of the game are independents, find
a)
P(Adam wins all 3 games)
b)
P(all 3 games end in a draw)
c)
P(Adam wins 1 and Beth wins 2 games)
d)
P(Adam and Beth win one game each)
A
B
A
D
A
B
B
D
D
A
B
D
A
a) 0.2 x 0.2 x 0.2 = 0.008
B
D
b) 0.45 x 0.45 x 0.45 = 0.091125
A
B
c) P(ABB)+P(BAB)+P(BBA)
D
A
B
(0.2 x 0.35 x 0.35) x 3
DA
0.0245 x 3 = 0.0735
B
D
A
B
d)
D
A
B P(ABD)+P(ADB)+P(BAD)+P(BDA)+P
A D (DAB)+P(DBA)
B
(0.2 x 0.35 x 0.45) x 6
DA
B
0.0315 x 6 = 0.189
AD
B
D
In a college there are 100 students taking A level French, German or
Spanish. Of these students, 64 are female and the rest are male.
There are 50 French students of whom 40 are female and 30 German
students of whom 10 are female.
Find the probability that a randomly chosen student
a)
Is taking Spanish
b) Is male, given that the student is taking Spanish
College records indicate that 70% of the French students, 80% of the
German students and 60% of the Spanish students have applied for
university.
A student is chosen at random
c) Find the probability that this student has applied for University
d) Given the student had applied for University, find the probability that
the student is studying French
Female
Male
Total
Applied to University
French
40
10
50
70% of 50 = 35
German
10
20
30
80% of 30 = 24
Spanish
14
6
20
60% of 20 = 12
Totals
64
36
100
Total = 71
a) P(Spanish) = 20/100 = 1/5
b) P(M|Spanish) = P(M  S) = 6/100 = 6/20= 3/10
20/
S
100
c) P(Applied to Uni) = 71/100
d) P(French|Uni) = P(F  U) = 35/100 = 35/71
71/
U
100
A researcher thinks there is a link between a person's height and level of
confidence. She measured the height h, to the nearest cm, of a random
sample of 9 people. She also devised a test to measure the level of
confidence c of each person. The data are shown in the table below
H
179
169
187
166
162
193
161
177
168
C
569
561
579
561
540
598
542
565
573
[You may use h2 = 272 094,
(a)
c2 = 2 878 966,
Find exact values of Shc Shh and Scc.
hc = 884 484]
(4)
(c)
Calculate the value of the product moment correlation coefficient for
these data.
(3)
(d)
Give an interpretation of your correlation coefficient. (1)
(b)
Shc = 884484 – 1562x5088 = 1433⅓
9
Shh = 10002/9; Scc = 2550
(c)
r=
1433 ⅓
√10002/9 x 2550
= 0.897488….
(d)
Taller people tend to be more confident
The number of caravans on Seaview caravan site on each night in August last
year is summarised in the following stem and leaf diagram.
Caravans
1
0
2
1
3
0
4
1
5
2
6
2
(a)
10 means 10
5
2
3
1
3
3
4
3
3
6
4
8
3 4 7 8 8
5 8 8 8 9 9
6 7
Totals
(2)
(4)
(8)
(9)
(5)
(3)
Find the three quartiles of these data.
(3)
During the same month, the least number of caravans on Northcliffe caravan site
was 31. The maximum number of caravans on this site on any night that month
was 72. The three quartiles for this site were 38, 45 and 52 respectively.
(b) On graph paper and using the same scale, draw box plots to represent the
data for both caravan sites. You may assume that there are no outliers.
(6)
(c) Compare and contrast these two box plots.
(3)
(d) Give an interpretation to the upper quartiles of these two distributions.(2)
a) Q1 = 33, Q2 = 41, Q3 = 52
B1B1B1
b) Median of Northcliffe is greater than median of Seaview
Upper quartiles are the same
IQR of Northcliffe is less than IQR of Seaview
Northcliffe positive skew, Seaview negative skew
Northcliffe symmetrical, Seaview positive skew (quartiles)
Range of Seaview greater than range of Northcliffe
B1 B1 B1 3
3
d) On 75% of the nights that month
both had no more than 52 caravans on site.
B1
B1
2
A discrete random variable has the following probability distribution
X
P(X=x)
Find
a) P(1 < X < 5)
b) P(2 ≤ X ≤ 4)
c) P(3 < X ≤ 6)
d) P(X < 3)
1
2
0.2 0.15
3
0.2
4
0.3
5
6
0.1 0.05
SOLUTION
A discrete random variable has the following probability distribution
X
P(X=x)
1
2
0.2 0.15
3
0.2
4
0.3
5
6
0.1 0.05
P(1 < X < 5) = P(X=2,3 or 4) = 0.15 + 0.2 + 0.3 = 0.65
P(2 ≤ X ≤ 4) = P(X=2,3 or 4) = 0.15 + 0.2 + 0.3 = 0.65
P(3 < X ≤ 6) = P(X=4,5 or 6) = 0.3 + 0.1 + 0.05 = 0.45
P(X < 3) = P(X=1 or 2) = 0.2 + 0.15 = 0.35
Eight students took tests in mathematics and physics. The marks for each student
are given in the table below where m represents the mathematics mark and p the
physics mark.
Students
Mark
A
B
C
D
E
F
G
H
M
9
14
13
10
7
8
20
17
p
11
23
21
15
19
10
31
26
A science teacher believes that students’ marks in physics depend upon their mathematical
ability. The teacher decides to investigate this relationship using the test marks.
(a)Draw a scatter diagram to illustrate these data.
(b)Showing your working, find the equation of the regression line of p on m.
(c)Draw the regression line on your scatter diagram.
(3)
(8)
(2)
A ninth student was absent for the physics test, but she sat the mathematics test and
scored 15.
(d)Using this model, estimate the mark she would have scored in the physics test. (2)
(a) scales and labels
B1
points
(6,7 points)
B2
B1
Line
M1 A1
(b) Sm = 98; Sp = 156; Sm2 = 1348; Smp = 2119
Smp = 2119 - 98 x 156 = 208
M1 A1
8
Smm = 1348 - 982 = 147.5
8
A1
 b = Smp = 208 = 1.410169 (awrt 1.41)
Smm 147.5
M1 A1
 a = 156 - (1.410169…) × 98 = 2.225429 (awrt 2.23)M1 A1
8
8
  p = 2.23 + 1.41m
(c) Line on graph
(d) p = 2.23 + 1.41 × 15 = 23.38
A1 ft
M1 A1
M1 A1