The table gives the distribution of heights in metres of 100 students.
| Height | 1.40-1.42 | 1.43-1.45 | 1.46-1.48 | 1.49-1.51 | 1.52-1.54 | 1.55-1.57 | 1.58-1.60 | 1.61-1.63 |
| Freq | 2 | 4 | 19 | 30 | 24 | 14 | 6 | 1 |
(a) Calculate the : (i) mean height ; (ii) mean deviation of the distribution.
(b) What is the probability that the height of a student selected at random is greater than the mean height of the distribution?
Explanation
| Height |
Class Mark (x) |
\(f\) | \(fx\) | \(d = x - \bar{x}\) | \(|d|\) | \(fd\) |
| 1.40-1.42 | 1.41 | 2 | 2.82 | -0.1 | 0.1 | 0.2 |
| 1.43-1.45 | 1.44 | 4 | 5.76 | -0.07 | 0.07 | 0.28 |
| 1.46-1.48 | 1.47 | 19 | 27.93 | -0.04 | 0.04 | 0.76 |
| 1.49-1.51 | 1.50 | 30 | 45.00 | -0.01 | 0.01 | 0.30 |
| 1.52-1.54 | 1.53 | 24 | 36.72 | 0.02 | 0.02 | 0.48 |
| 1.55-1.57 | 1.56 | 14 | 21.84 | 0.05 | 0.05 | 0.70 |
| 1.58-1.60 | 1.59 | 6 | 9.54 | 0.08 | 0.08 | 0.48 |
| 1.61-1.63 | 1.62 | 1 | 1.62 | 0.11 | 0.11 | 0.11 |
| 100 | 151.23 | 3.31 |
(a)(i) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{151.23}{100}\)
\(\approxeq 1.51cm\)
(ii) Mean deviation = \(\frac{\sum fd}{\sum f}\)
= \(\frac{3.31}{100}\)
= \(0.033\)
(b) p(height is gretaer than mean height) = \(\frac{24 + 14 + 6 + 1}{100}\)
= \(0.45\)