| Age | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No. of children | 2 | 6 | 5 | x | 6 | 9 | 8 | 5 |
The table shows the distribution of ages of a number of children in a school. If the mean of the distribution is 7, find the;
(a) value of x, (b) standard deviation of their ages.
Mean = \(\frac{\sum fx}{ \sum f}\) = (3x2) +(4x6) + (5x5) + (6x) + (7x6) + (8x9) + (9x8) + (10x5) ÷ (2 + 6 + 5 + x + 6 + 9 + 8 + 5)
7 = (291 + 6x) ÷ (41+x)
7(41+x) = 291 + 6x
287 + 7x = 291 + 6x
x = 4
(b)
| Age(x in yrs) | No.of child | Fx\(^2\) |
| 3 | 2 | 18 |
| 4 | 6 | 96 |
| 5 | 5 | 125 |
| 6 | 4 | 144 |
| 7 | 6 | 294 |
| 8 | 9 | 576 |
| 9 | 8 | 648 |
| 10 | 5 | 500 |
| Σf = 45 |
Σfx\(^2\) = 2401 |
Standard deviation = \(\sqrt \frac{196}{45} = \sqrt 4.3555\)
= 2.087
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