The weights to the nearest kilogram, of a group of 50 students in a College

MATHEMATICS
WAEC 1990

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.

Explanation

(a)
Class IntervalTallyFreq
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 - 4947-152256-901350
50 - 5452-101009-90900
55 - 5957-52510-50250
60 - 6462001200
65 - 6967525735175
70 - 747210100660600
\(\sum\)50-1353275
\(Mean (\bar{x}) = A + \frac{\sum f(x - A)}{\sum f}\) = \(62 + \frac{-135}{50}\) = \(62 - 2.7 = 49.3\) 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)


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