Question

The Whitcomb Company manufactures a metal ring for industrial engines that usually weighs

about 50 ounces. A random sample of 40 of these metal rings produced the following weights
(in ounces).
51 53 50 56 50 47 44 51
53 44 52 41 57 51 55 38
41 39 51 46 63 52 43 49
44 44 54 69 42 53 52 50
57 43 50 50 47 36 43 62
a) Calculate the number of classes needed to construct group distribution. (3 marks)
b) Calculate the class size (class width). (3 marks)
c) Construct a frequency distribution table for the data given above. (14 marks)
Time (min)
Interval
Frequency Cumulative
Frequency
Percentage Cumulative
frequency

Answer 1

a) The number of classes needed to construct group distribution is 6.

b) The class size (class width) should be 6.

c) The frequency distribution table for the data give above is as follows:

Interval Frequency Cumulative Percentage Cumulative

Frequency Percentage

36-41 5 5 12.5% 12.5%

42-47 8 13 20% 32.5%

48-53 14 27 35% 67.5%

54-59 7 34 17.5% 85%

60-65 4 38 10% 95%

66-71 2 40 5% 100%

a) The number of classes (or bins) in a frequency distribution is often chosen using Sturges’ rule, which is given by:

k = 1 + 3.322log₁₀​(n)

where:

k = the number of classes

n = the number of observations.

In this case, n = 40, the formula gives:

k = 1 + 3.322log₁₀​(40) ≈ 6

Thus, we need approximately 6 classes to construct the frequency distribution.

b) The class size (or class width) is calculated as follows:

w =
`(R)/(k)`​

where:

w = the class width,

R = the range of the data (i.e., the difference between the maximum and minimum values), and

k = the number of classes.

The range of the data is 69−36=33, so substituting this and the number of classes into the formula gives:

w =
`(33)/(6)`​ = 5.5 ≈ 6

c) Constructing the frequency distribution table with the minimum value as 36 and the class size as 6, the class intervals will be 36-41, 42-47, 48-53, 54-59, 60-65, and 66-71: