The table shows the distribution of ages of 22 students in a school.
| Age (years) | 12-14 | 15-17 | 18-20 | 21-23 | 24-26 |
| Frequency | 6 | 10 | 3 | 2 | 1 |
Using an assumed mean of 19, calculate, correct to three significant figures, the :
(a) mean age ; (b) standard deviation ; of the distribution.
Explanation
Assumed mean, A = 19.
| Age (years) | Mid-age (x) | Frequency | \(d = x - A\) | \(fd\) | \(fd^{2}\) |
| 12 - 14 | 13 | 6 | -6 | -36 | 216 |
| 15 - 17 | 16 | 10 | -3 | -30 | 90 |
| 18 - 20 | 19 | 3 | 0 | 0 | 0 |
| 21 - 23 | 22 | 2 | 3 | 6 | 18 |
| 24 - 26 | 25 | 1 | 6 | 6 | 36 |
| 22 | -54 | 360 |
(a) \(\bar{x} = A + \frac{\sum fd}{\sum f}\)
= \(19 + \frac{-54}{22}\)
= \(19 - 2.455\)
= \(16.545 \approxeq 16.5\) years.
(b) Standard deviation , \(\sigma = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd}{\sum f})}\)
= \(\sqrt{\frac{360}{22} - (\frac{-54}{22})}\)
= \(\sqrt{16.364 - (2.45)^{2}}\)
= \(\sqrt{16.364 - 6.025}\)
= \(\sqrt{10.339}\)
= 3.215 \(\approxeq\) 3.22 years.