Handling Data 2

5 marks · Hard

The table shows the results of a survey into how long 60 customers spent in a shop.

(a)

Estimate the percentage of customers who spent at least 20 minutes in the shop.

[3 marks]

(b)

Find the modal class of the time spent.

[1 mark]

(c)

Calculate an estimate for the mean time spent in the shop. Give your answer to 2 decimal places.

[1 mark]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

Show answer

(a) 62%

(b) 20 ≤ t < 30

Common mistakes: Not including 20 minute boundary in calculation; Not using midpoints; Arithmetic errors

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Identifies correct classes for 'at least 20 minutes'.

M2 (a) · 1 mark · Adds the relevant frequencies (22+11+4). (follow-through)

A1 (a) · 1 mark · Divides by 60 and multiplies by 100 for percentage; rounds appropriately. (follow-through)

B1 (b) · 1 mark · Correctly states modal class as 20 ≤ t < 30.

A1 (c) · 1 mark · Calculates grouped mean using correct midpoints and gives answer to 2 dp.

Worked solution

a) Classes for at least 20 minutes: 20 ≤ t < 30, 30 ≤ t < 40, 40 ≤ t ≤ 50.

Frequencies: 22 + 11 + 4 = 37

Percentage = (37 / 60) × 100 = 61.666... ≈ 62%

b) Modal class: Highest frequency is 22, so 20 ≤ t < 30.

c) Midpoints:

Class	Midpoint	Frequency (f)	f × x
0 ≤ t < 10	5	9	45
10 ≤ t < 20	15	14	210
20 ≤ t < 30	25	22	550
30 ≤ t < 40	35	11	385
40 ≤ t ≤ 50	45	4	180

Sum of f × x: 45 + 210 + 550 + 385 + 180 = 1370

Total frequency = 60