The table shows the distribution of masks obtained by students in an examination.
| Marks | 50 – 54 | 55 – 59 | 60 – 64 | 65 – 69 | 70 – 74 | 75 – 79 | 80 – 84 | 85 – 89 |
| Frequency | 5 | 15 | 20 | 28 | 12 | 9 | 7 | 4 |
Using an assumed mean of 67, calculate, correct to one decimal place. the
a) Mean
b) Standard deviation of the distribution
Explanation
| x | f | d | fd | (delta) | f(delta) |
| 52 | 5 | -15 | -75 | 225 | 1125 |
| 57 | 15 | -10 | -150 | 100 | 1500 |
| 62 | 20 | -5 | -100 | 25 | 500 |
| 67 | 28 | 0 | 0 | 0 | 0 |
| 72 | 12 | 5 | 60 | 25 | 300 |
| 77 | 9 | 10 | 90 | 100 | 900 |
| 82 | 7 | 15 | 105 | 225 | 1575 |
| 87 | 4 | 20 | 80 | 400 | 1600 |
| Total | 100 | 10 | 7500 |
(a) To find mean; = 67 + (frac{10}{100}) = 67.1
(b) Substitute and find the standard deviation as (sqrt{frac{7500}{100} – (frac{10}{100})^2}) = (sqrt{74.99})
= 8.6597
= 8.7, correct to one decimal place