# CIE A level Math -Probability & Statistics 1 : 5.1 Representation of data: cumulative frequency graph: 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

Box A contains 7 red balls and 1 blue ball. Box B contains 9 red balls and 5 blue balls. A ball is chosen at random from box A and placed in box B. A ball is then chosen at random from box B. The tree diagram below shows the possibilities for the colours of the balls chosen.
(a) Complete the tree diagram to show the probabilities.

(b) Find the probability that the two balls chosen are not the same colour.
(c) Find the probability that the ball chosen form box A is blue given that given that the ball chosen form box B is blue

Ans:

(a)

(b) $$\frac{7}{8} \times \frac{5}{15}+\frac{1}{8} \times \frac{9}{15}$$
$$=\frac{44}{120}[\frac{11}{30}or0.367]$$
(c) P(A blue|B blue) $$=\frac{A blue \bigcap B blue)}{P(B blue)}$$
$$=\frac{\frac{1}{8}\times \frac{6}{15}}{\frac{7}{8}\times \frac{5}{15}+\frac{1}{8}\times \frac{6}{15}}=\frac{\frac{1}{20}}{\frac{41}{120}}$$
$$=\frac{6}{41}$$ or 0.146

### Question

A driver records the distance travelled i each of 150 journeys. These distances, correct to the nearest km, are summarised in the following table.

 Distance(km) 0-4 5-10 11-20 21-30 31-40 41-60 Frequency 12 16 32 66 20 4

(a) Draw a cumulative frequency graph to illustrate the data.

(b) For 30% of these journeys the distance travelled is d km or more.
Use your graph to estimate the value of d.
(c) Calculate an estimate of the mean distance travelled for the 150 journeys.

Ans:

(a)

 Distance 0-4 5-10 11-20 21-30 31-40 41-60 Upper boundary 4-5 10-5 20-5 30-5 40.5 60-5 Cumulative frequency 12 28 60 126 146 150

(b) 70% of 150 = 105
Approx. 27

(c) Midpoints: 2.25, 7.5, 15.5, 25.5, 35.5, 50.5
Mean = $$\frac{2.25 \times 12 + 7.5 \times 16+ + 15.5 \times 32 + 25.5 \times 66 + 65.5 \times 20 + 50.5 \times 4}{150}$$
$$=\frac{27+120+496+1686+710+202}{150}$$
$$[=\frac{3238}{150}]=21.6, 21\frac{44}{75}$$

### 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.                                                                                                  [5]

(b) What is the greatest possible value of the interquartile range for the data?                                                    [2]

(c) Calculate estimates of the mean and standard deviation of the number of chocolate bars sold.                 [4]

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)

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