# CIE A level Math -Probability & Statistics 1 : 5.1 Representation of data: measures of central tendency and variation : Exam Style Questions Paper 5

At a summer camp an arithmetic test is taken by 250 children. The times taken, to the nearest minute, to complete the test were recorded. The results are summarised in the table.

 Time taken, in minutes 1-30 31-45 46-65 66-75 76-100 Frequency 21 30 68 86 45

### (a) Question

Draw a histogram to represent this information. Ans:

 Class Width 30 15 20 10 25 Frequency Density 0.7 2 3.4 8.6 1.8

### (b) Question

State which class interval contains the median.

Ans:

66 – 75

### (c) Question

Given that an estimate of the mean time is 61.05 minutes, state what feature of the distribution accounts for the median and the mean being different.

Ans:

Distribution is not symmetrical

### Question

The random variable X takes the values 1, 2, 3, 4 only. The probability that X takes the value x is kx(5-x), where k is a constant.

(a) Draw up the probability distribution table for X, in terms of k.

(b) Show that Var(X)=1.05.

Ans:

(a)

 x 1 2 3 4 prob 4k 6k 6k 4k

(b) $$[4k+6k+6k+4k=1]k=\frac{1}{20}(=0.05)$$
$$E(X)=1\times \frac{4}{20}+2 \times \frac{6}{20}+3\times \frac{6}{20}+4\times \frac{4}{20}=\frac{4}{20}+\frac{12}{20}+\frac{18}{20}+\frac{16}{20}$$
$$Var(X)=1^2\times \frac{4}{20}+2^2\times \frac{6}{20}+3^2\times \frac{6}{20}+4^2 \times \frac{4}{20}-(their2\frac{1}{2})^2$$
$$=(4+24+54+64)\times their 0.05-(their2.5)^2$$
Or
$$(1-2.5)^2\times \frac{4}{20}+(2-2.5)^2\times \frac{6}{20}+(3-2.5)^2\times \frac{6}{20}+(4-2.5)^2\times \frac{4}{20}$$
1.05

### Question 4.

4 A survey was made of the journey times of 63 people who cycle to work in a certain town. The results are summarised in the following cumulative frequency table. (i) State how many journey times were between 25 and 45 minutes. 
(ii) Draw a histogram on graph paper to represent the data. 
(iii) Calculate an estimate of the mean journey time. 

Ans:  ### Question

Helen measures the lengths of 150 fish of a certain species in a large pond. These lengths, correct to the nearest centimetre, are summarised in the following table.

 Length(cm) 0-9 10-14 15-19 20-30 Frequency 15 48 66 21

(a) Draw a cumulative frequency graph to illustrate the data.
(b) 40% of these fish have of d cm or more. Use your graph to estimate the value of d.
The mean length of these 150 fish is 15.295 cm.
(c) Calculate an estimate for the variance of the lengths of the fish.

Ans:

(a) (b) 60% of 150 = 90
Approx. 16.5 [cm]
(c) Midpoints: 4.75, 12, 17, 25
Var = $$4.75^2 \times 15 + 12^2 \times 48 + 17^2 \times 66 + 25^2 \times 21}{150}-15.295^2$$
= 29.1

### Question

For 10 values of x the mean is 86.2 and Σ(x − a) = 362. Find the value of

(i) Σ x, 

(ii) the constant a. 

Ans:

1 (i) Σx = 862

(ii) 362/10 + a = 86.2
a = 50

### Question

The numbers of chocolate bars sold per day in a cinema over a period of 100 days are summarised in
the following table. (a) Draw a histogram to represent this information. (b) What is the greatest possible value of the interquartile range for the data?                                                    

(c) Calculate estimates of the mean and standard deviation of the number of chocolate bars sold.                 

Ans

7 (a) Class widths: 10, 5, 15, 20, 10
Frequency density = frequency/their class width: 1.8, 4.8, 2, 1, 0.8
All heights correct on diagram (using a linear scale)
Correct bar ends
Bar ends: 10.5, 15.5, 30.5, 50.5, 60.5

7 (b) 11 – 15 and 31 – 50 B1
Greatest IQR = 50 – 11 = 39

7 (c) $$Mean=\frac{18\times 5.5+24\times 13+30\times 23+20\times 40.5+8\times 55.5}{100}=\frac{2355}{100}=23.6$$

$$Var=\frac{18\times 5.5^{2}+24\times 13^{2}+30\times 23^{2}+20\times 40.5^{2}+8\times 55.5^{2}}{100}-mean^{2}$$

$$\frac{77917.5}{100}-mean^{2}=224.57$$

Standard deviation = 15.0
(FT their variance)

### Question

Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line.
(a) Find the number of different arrangements of the 11 candles if there is a red candle at each end.
(b) Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together.

Ans:

(a) R^^^^^^^^^R
$$\frac{9!}{3!6!}$$
= 84
(b) ^(BBB)^^^^^
$$\frac{7!}{6!} \times \frac{8 \times 7}{2}$$
= 196
Alternative for question 4(b)
[Arrangements, blue together – Arrangements with blues together and reds together=]
$$\frac{9!}{2!6!}-\frac{8!}{6!}\} = [252 – 56] = 196 A summary of 40 values of x gives the following information: Σ (x – k) = 520, Σ ( x – k)2 = 9640, where k is a constant. ### (a) Question Given that the mean of these 40 values of x is 34, find the value of k. Answer/Explanation Ans: \(\left [ \frac{\sum x}{40} – k = \frac{\sum (x-k)}{40}\right ]$$

$$\frac{40 \times 34}{40} – k = \frac{520}{40}$$

k [= 34 – 13] = 21

### (b) Question

Find the variance of these 40 values of x.

Ans:

$$Var = \left [ \frac{\sum (x – k)^{2}}{40} – (\frac{\sum (x – k)^{}}{40})^{2} \right ] = \frac{9640}{40} – \left ( \frac{520}{40} \right )^{2} = [241 – 13^{2} =]$$

72

The times, in minutes, that Karli spends each day on social media are normally distributed with mean 125 and standard deviation 24.

### (a) (i) Question

On how many days of the year (365 days) would you expect Karli to spend more than 142 minutes on social media?

Ans:

$$P(X>142) = P\left ( Z>\frac{142-125}{24} \right )$$

[=P(Z>0.7083) = ] 1-0.7604

0.2396

Their 0.2396 × 365 [=87.454]

87 or 88

### (ii) Question

Find the probability that Karli spends more than 142 minutes on social media on fewer than 2 of 10 randomly chosen days.

Ans:

P(0, 1) = 0.760410 + 10C1 × 0.23961  × 0.76049

[ = 0.064628 + 0.20364]

0.268

### (b) Question

On 90% of days, Karli spends more than t minutes on social media.
Find the value of t.

Ans:

z = ±1.282

$$\frac{t – 125}{24} = -1.282$$

t = 94.2

### Question

The times, t minutes, taken by 150 students to complete a particular challenge are summarised in the
following cumulative frequency table. (a) Draw a cumulative frequency graph to illustrate the data. (b) 24% of the students take k minutes or longer to complete the challenge. Use your graph to
estimate the value of k.                                                                                                                                    

(c) Calculate estimates of the mean and the standard deviation of the time taken to complete the
challenge.                                                                                                                                                           

Ans

6 (a) Correct cumulative frequency curve

6  (b) 150× 0·76 = 114
k = 45 (mins)

6 (c) Frequencies: 12 36 58 28 16

$$Mean=\frac{10\times 12 \times 25 \times 36\times 35\times 58 \times 50 \times 28\times 80\times 16}{150}$$

$$\frac{120 + 900 + 2030+ 1400 + 1280}{150}$$

$$38.2, 38\frac{1}{5}$$

$$Variance=\frac{12\times 10^{2}\times 36\times 25^{2}\times 58\times 35^{2}\times 28\times 50^{2}\times 16\times 80^{2}}{150}-mean^{2}$$

$$=\frac{1200 + 22500 + 71050 + 70000 + 102400}{150}-mean^{2}$$

(Standard deviation $$=\sqrt{321.76}=17.9$$

### Question

The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.
(a) Show that the probability that the score is 4 is $$\frac{1}{12}$$                                                                

The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is
denoted by the random variable X.
(b) Find the mean of X.                                                                                                                                             

(c) Find the probability that a score of 4 is first obtained on the 6th throw.                                                 

(d) Find P(X < 8).                                                                                                                                                       

1 (a) Prob of 4 (from 1,3, 3,1 or 2,2) $$=\frac{3}{36}=\frac{1}{12} \ AG$$
1 (b) $$Mean=\frac{1}{\frac{1}{12}}=12$$
1 (c) $$\left ( \frac{11}{12} \right )\times \frac{1}{12}=0.0539\ or \frac{161051}{2985984}$$
1 (d) $$1-\left ( \frac{11}{12} \right )^{7}$$