1. 1. A survey conducted about job satisfaction showed that 20 % of workers are not happy with their
current jobs. Assume that this result is true for the population of all workers. Two workers are
selected at random, and asked whether or not they are happy with their current jobs. Find the
probability that,
(a) both are not happy with their current jobs, (4 marks)
Probability 2 people are not happy is 1/5 × 1/5 = 1/25.
(b) at least one of them is happy with his current job, (6 marks)
Combinations are HU, UH, HH.
4/5 × 1/5 + 1/5 × 4/5 + 4/5 × 4/5 = 24/25
(c) less than two are happy with their current jobs. (6 marks)
Combinations are HU, UH, UU.
4/5 × 1/5 + 1/5 × 4/5 + 1/5 × 1/5 = 9/25
2. A triangle ABC has a right angle at B. Side AB measures 7.5 cm and side BC measures 10.5 cm.
Giving your answers correct to the nearest 1 decimal place,
(a) Find the length of side AC. (6 marks)
AC =
√
7.52 + 10.52 =
√
166.5 = 12.9 to 1 d.p.
(b) Find the value of angle C. (6 marks)
AB = opp, BC = adj.
tan(C) = 7.5
10.5 = 0.714
C = tan−1
(0.714) = 35.5◦
to 1 d.p.
3. The turnout of spectators at a motor rally is dependent upon the weather. The weather forecast
gives a probability of 0.75 that it will rain on the day of the race. On a rainy day the probability
of a big turnout is 0.4, but if it does not rain, the probability of a big turnout increases to 0.9.
(a) Draw a tree diagram to represent this information. (3 marks)
Rain
0.75
No Rain
0.25
Big turnout0.4
Small turnout0.6
Big turnout0.9
Small turnout0.1
(b) Find the probability that there will be a big turnout and it rains. (3 marks)
0.75 × 0.4 = 0.3.
(c) Find the probability that there will be a big turnout. (6 marks)
0.75 × 0.4 + 0.25 × 0.9 = 0.525
1

2. 4. In a study of how students use their mobile telephones, the phone usage of a random sample of
11 students was examined for a particular week. The total length of calls in minutes, for the 11
students were
53 51 35 23 36 17 77 54 55 60 110
(a) Find the median length of calls. (4 marks)
The ordered table is
17 23 35 36 51 53 54 55 60 77 110
11 × 1/2 = 5.5
The median is the 6th
term 53.
(b) Calculate the interquartile range. (6 marks)
11 × 1/4 = 2.75
The first quartile Q1 is the 3rd
term 35.
11 × 3/4 = 8.25
The third quartile Q3 is the 9th
term 60.
The interquartile range is Q3 − Q1 = 60 − 35 = 25.
5. Triangle ABC has AB = 6 cm, AC = 5 cm and BC = 4 cm . Giving your answers correct to 2
decimal places,
A B
C
5 4
6
(a) Show that cosA = 3⁄4. (6 marks)
cos(A) = b2
+c2
−a2
2bc = 52
+62
−42
2×5×6 = 0.75
A = cos−1
(0.75) = 41.4 to 1 d.p.
(b) Hence, or otherwise, find angle C correct to 1 decimal place. (6 marks)
c
sin(C) = a
sin(A)
6
sin(C) = 4
sin(41.4)
sin(C) = 6×sin(41.4)
4 = 0.99
C = sin−1
(0.99) = 82.7.
(c) Find the area of the triangle ABC. (6 marks)
Area = 0.5 × 5 × 6 × sin(41.4) = 9.92 to 2d.p.
2

3. 6. There are two routes for an undergraduate student to walk to the University campus. She recorded
the time it takes over a series of six journeys for each route and results are shown in the table
below.
Route 1 21 16 10 12 14 11
Route 2 9 14 17 15 16 13
(a) Work out the mean time taken for each route. (6 marks)
¯x1 = 21 + 16 + 10 + 12 + 14 + 11/6 = 14
¯x2 = 9 + 14 + 17 + 15 + 16 + 13/6 = 14
(b) Calculate the variance and standard deviation of each of the two routes. (8 marks)
σ2
=
x2
n − ¯x2
The squared data is
441 256 100 144 196 121
81 196 289 225 256 169
σ2
1 =
6
i=1
1[i]2
6 − 142
=441+256+100+144+196+121⁄6-196=41⁄3
σ1 = 3.7 to 1 d.p.
σ2
2 =
6
i=1
1[i]2
6 − 142
=81+196+289+225+256+169⁄6-196=20⁄3
σ2 = 2.6 to 1 d.p.
(c) Using your answers to a and b, suggest which route you would recommend . State your
reason clearly. (2 marks)
The average travel time for both journeys is the same. I would choose the second route
as the standard deviation is smaller so there is a smaller chance of being late.
7. Zhou is getting married and wishes to rent out his one bedroom apartment. A property manage-
ment advisor tells him that the average monthly rent for a one bedroom apartment in his area is
244. To test this, he approaches a property rental agency and selects 7 one bedroom apartments
at random from those available for rent. The monthly rentals of these are;
310 290 215 270 220 245 200
Test at 5 % level of significance if the advisor’s claim is different from the average rent of 244.
(16 marks)
µ = 244. Only told a mean so will be doing a T-test.
H0 : µ = 244
H1 : µ = 244
We are doing a 2 tailed test at a 5% level of significance so we are looking for a probability of
0.05 × 1/2 = 0.025.
Our degree of freedom is 7 − 1 = 6 so our critical value is 2.45.
To calculate the test statistic we need the sample mean, ¯x and the sample standard deviation
σ(X).
¯x = 310 + 290 + 215 + 270 + 220 + 245 + 200/7 = 250.
The sample variance is given by
s2
=
x2
− 7 × ¯x2
6
The squared data is
96100 84100 46225 72900 48400 60025 40000
The sample variance is
s2
=
96100 + 84100 + 46225 + 72900 + 48400 + 60025 + 40000 − 7 × µ2
(X)
6
=
5125
3
3

4. The test statistic is
T =
¯x − A
s/
√
n
=
250 − 244
41.332/
√
7
= 0.38 to 2 d.p.
We have −2.45 < 0.38 < 2.45 so we accept the null hypothesis.
4