The weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:
65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.
(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.
(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a)
Class Interval
Tally
Freq
45 - 49
|||| |
6
50 - 54
|||| ||||
9
55 - 59
|||| ||||
10
60 - 64
|||| |||| ||
12
65 - 69
|||| ||
7
70 - 74
|||| |
6
(b)
Class
Interval
Mid-value
(x)
x - 62
\((x - 62)^{2}\)
\(f\)
\(f(x - 62)\)
\(f(x - 62)^{2}\)
45 - 49
47
-15
225
6
-90
1350
50 - 54
52
-10
100
9
-90
900
55 - 59
57
-5
25
10
-50
250
60 - 64
62
0
0
12
0
0
65 - 69
67
5
25
7
35
175
70 - 74
72
10
100
6
60
600
\(\sum\)
50
-135
3275
\(Mean (\bar{x}) = A + \frac{\sum f(x - A)}{\sum f}\)
= \(62 + \frac{-135}{50}\)
= \(62 - 2.7 = 59.3\)
Standard deviation = \(\sqrt{\frac{\sum f(x - A)}{\sum f}}\)
= \(\sqrt{\frac{3275}{50}}\)
= \(\sqrt{65.5}\)
= \(8.093 \approxeq 8.1\) (to 1 decimal place)
Source: Myschool Past Questions Archive (Record #54646)