Handling Data 2

5 marks · Hard

The frequency table shows the lengths, in cm, of a sample of 40 pencils measured in a classroom.

Lengths of pencils
Length (L, cm)	Frequency
10 ≤ L < 12	5
12 ≤ L < 14	12
14 ≤ L < 16	17
16 ≤ L ≤ 18	6

(a)

Estimate the mean length of the pencils. Give your answer to 2 decimal places.

[3 marks]

(b)

State the modal class.

[1 mark]

(c)

Find the class interval that contains the median length.

[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) 14.18 cm

(b) 14 ≤ L < 16

Common mistakes: Using endpoints instead of midpoints; Not dividing by total frequency; Choosing wrong modal class

Show mark scheme and worked solution

Marking points

M1 (a) · 1 mark · Uses midpoints for each group and calculates f × x values.

M2 (a) · 1 mark · Adds fx and divides by total frequency. (follow-through)

A1 (a) · 1 mark · Estimate to 2 decimal places correctly (accept 14.17 or 14.18). (follow-through)

B1 (b) · 1 mark · Correctly identifies the modal class (highest frequency).

B1 (c) · 1 mark · Correctly finds the median class (contains the 20th/21st value).

Worked solution

a) Find midpoints of each class:

Class	Midpoint (x)	Frequency (f)	f × x
10 ≤ L < 12	11	5	55
12 ≤ L < 14	13	12	156
14 ≤ L < 16	15	17	255
16 ≤ L ≤ 18	17	6	102

Sum of frequencies = 5 + 12 + 17 + 6 = 40

Sum of f × x = 55 + 156 + 255 + 102 = 568

(Or 14.18 to 2 decimal places)

b) Modal class is the one with highest frequency: 17 (for 14 ≤ L < 16)

c) Median class: total frequency = 40, so median position is (40+1)/2 = 20.5th item; add frequencies cumulatively: Class 1: 5 Class 2: 5 + 12 = 17 Class 3: 17 + 17 = 34 20.5th item falls in third class: 14 ≤ L < 16