The table shows the distribution of masks obtained by students in an examination. Marks 50 - 54 55 - 59 60 - 64 65 - 69 70 - 74 75 - 79 80 - 84 85 - 89 Freq...
Question 1 Report
The table shows the distribution of masks obtained by students in an examination.
| Marks | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | 75 - 79 | 80 - 84 | 85 - 89 |
| Frequency | 5 | 15 | 20 | 28 | 12 | 9 | 7 | 4 |
Using an assumed mean of 67, calculate, correct to one decimal place. the
a) Mean
b) Standard deviation of the distribution
Answer Details
Method (assumed mean / coding). Take assumed mean \(A = 67\) and class width \(c = 5\). For each class let \(x\) be the midpoint and \(u = \dfrac{x - A}{c}\).
| Marks | Midpoint \(x\) | \(u\) | \(f\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 50 - 54 | 52 | -3 | 5 | -15 | 45 |
| 55 - 59 | 57 | -2 | 15 | -30 | 60 |
| 60 - 64 | 62 | -1 | 20 | -20 | 20 |
| 65 - 69 | 67 | 0 | 28 | 0 | 0 |
| 70 - 74 | 72 | 1 | 12 | 12 | 12 |
| 75 - 79 | 77 | 2 | 9 | 18 | 36 |
| 80 - 84 | 82 | 3 | 7 | 21 | 63 |
| 85 - 89 | 87 | 4 | 4 | 16 | 64 |
| Total | 100 | 2 | 300 |
(a) Mean.
\[ \bar{x} = A + \left(\frac{\sum fu}{\sum f}\right)c = 67 + \frac{2}{100}\times 5 = 67 + 0.1 = \mathbf{67.1} \](b) Standard deviation.
\[ \text{SD} = c\sqrt{\frac{\sum fu^2}{\sum f} - \left(\frac{\sum fu}{\sum f}\right)^2} = 5\sqrt{\frac{300}{100} - \left(\frac{2}{100}\right)^2} \] \[ = 5\sqrt{3 - 0.0004} = 5\sqrt{2.9996} = 5 \times 1.7319 \approx \mathbf{8.7} \]Download The App On Google Playstore
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