In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table
| Number of absentees | 0-4 | 5-9 | 10-14 | 15-19 | 20-24 |
| Number of Days | 1 | 5 | 10 | 9 | 5 |
Calculate the : (a) Mean
(b) Standard deviation , correct to two decimal places.
Explanation
| \(x\) | mid-value | \(f\) | \(fx\) | \((x - \bar{x})\) | \((x - \bar{x})^{2}\) | \(f(x - \bar{x})^{2}\) |
| 0 - 4 | 2 | 1 | 2 | -12 | 144 | 144 |
| 5 - 9 | 7 | 5 | 35 | -7 | 49 | 245 |
| 10 - 14 | 12 | 10 | 120 | -2 | 4 | 40 |
| 15 - 19 | 17 | 9 | 153 | 3 | 9 | 81 |
| 20 - 24 | 22 | 5 | 110 | 8 | 64 | 320 |
| \(\sum \) | 30 | 420 | 830 |
(a) Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{420}{30} = 14\)
(b) Standard deviation = \(\sqrt{\frac{\sum f(x - \bar{x})^{2}}{\sum f}\)
= \(\sqrt{\frac{830}{30}}\)
= \(\sqrt{27.66}\)
= 5.26.