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.