The table shows the distribution of masks obtained by students in an examination.
Using an assumed mean of 67, calculate, correct to one decimal place. the
a) Mean
b) Standard deviation of the distribution
| 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
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Correct Answer: Option
Explanation:
(a) To find mean; = 67 + \(\frac{10}{100}\) = 67.1
(b) Substitute and find the standard deviation as \(\sqrt{\frac{7500}{100} - (\frac{10}{100})^2}\) = \(\sqrt{74.99}\)
= 8.6597
= 8.7, correct to one decimal place
| x | f | d | fd | \(\delta\) | f\(\delta\) |
| 52 | 5 | -15 | -75 | 225 | 1125 |
| 57 | 15 | -10 | -150 | 100 | 1500 |
| 62 | 20 | -5 | -100 | 25 | 500 |
| 67 | 28 | 0 | 0 | 0 | 0 |
| 72 | 12 | 5 | 60 | 25 | 300 |
| 77 | 9 | 10 | 90 | 100 | 900 |
| 82 | 7 | 15 | 105 | 225 | 1575 |
| 87 | 4 | 20 | 80 | 400 | 1600 |
| Total | 100 | 10 | 7500 |
(a) To find mean; = 67 + \(\frac{10}{100}\) = 67.1
(b) Substitute and find the standard deviation as \(\sqrt{\frac{7500}{100} - (\frac{10}{100})^2}\) = \(\sqrt{74.99}\)
= 8.6597
= 8.7, correct to one decimal place