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 1.61-1.6...
Question 1 Report
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?
Answer Details
Each class has width 3; the mid-values are \(1.41,1.44,\dots,1.62\). Use a working-mean coding \(u=\dfrac{x-A}{c}\) with \(A=1.50\) and \(c=0.03\).
| Height | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 1.40-1.42 | 1.41 | 2 | -3 | -6 | 18 |
| 1.43-1.45 | 1.44 | 4 | -2 | -8 | 16 |
| 1.46-1.48 | 1.47 | 19 | -1 | -19 | 19 |
| 1.49-1.51 | 1.50 | 30 | 0 | 0 | 0 |
| 1.52-1.54 | 1.53 | 24 | 1 | 24 | 24 |
| 1.55-1.57 | 1.56 | 14 | 2 | 28 | 56 |
| 1.58-1.60 | 1.59 | 6 | 3 | 18 | 54 |
| 1.61-1.63 | 1.62 | 1 | 4 | 4 | 16 |
| Total | 100 | 41 | 203 |
(a)(i) Mean height.\[\bar{x}=A+\frac{\sum fu}{\sum f}\,c=1.50+\frac{41}{100}(0.03)=1.50+0.0123=1.5123\approx 1.51\text{ m}\]
(a)(ii) Mean deviation. \(\text{MD}=\dfrac{\sum f\,|x-\bar{x}|}{\sum f}\) with \(\bar{x}=1.5123\).
| \(x\) | \(f\) | \(|x-\bar{x}|\) | \(f|x-\bar{x}|\) |
|---|---|---|---|
| 1.41 | 2 | 0.1023 | 0.2046 |
| 1.44 | 4 | 0.0723 | 0.2892 |
| 1.47 | 19 | 0.0423 | 0.8037 |
| 1.50 | 30 | 0.0123 | 0.3690 |
| 1.53 | 24 | 0.0177 | 0.4248 |
| 1.56 | 14 | 0.0477 | 0.6678 |
| 1.59 | 6 | 0.0777 | 0.4662 |
| 1.62 | 1 | 0.1077 | 0.1077 |
| Total | 100 | 3.333 |
\[\text{MD}=\frac{3.333}{100}=0.0333\approx 0.03\text{ m}\]
(b) Probability of height greater than the mean. The mean \(1.5123\) m lies below the upper boundary \(1.515\) m of the class \(1.49-1.51\), so every student in the classes \(1.52-1.54\) and above exceeds the mean:\[24+14+6+1=45\]\[P(\text{height}>\bar{x})=\frac{45}{100}=0.45\]
Mean = 1.51 m, mean deviation \(\approx\) 0.03 m, and the required probability = 0.45.
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