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 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 = 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)