| ClassInterval | Frequency |
| 60 - 64 | 2 |
| 65 - 69 | 3 |
| 70 - 74 | 6 |
| 75 - 79 | 11 |
| 80 - 84 | 8 |
| 85 - 89 | 7 |
| 90 - 94 | 2 |
| 95 - 99 | 1 |
The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the
(a) mean ; (b) standard deviation of the distribution.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{3150}{40}\)
= 78.75
(b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\)
= \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\)
= \(\sqrt{6264.75 - (78.75)^{2}}\)
= \(\sqrt{6264.75 - 6201.5625}\)
= \(\sqrt{63.1875}\)
= 7.95.
| ClassInterval | Freq(f) | Mid-point(x) | \(fx\) | \(x^{2}\) | \(fx^{2}\) |
| 60 - 64 | 2 | 62 | 124 | 3844 | 7688 |
| 65 - 69 | 3 | 67 | 201 | 4489 | 13487 |
| 70 - 74 | 6 | 72 | 432 | 5184 | 31104 |
| 75 - 79 | 11 | 77 | 847 | 5929 | 65219 |
| 80 - 84 | 8 | 82 | 656 | 6724 | 53792 |
| 85 - 89 | 7 | 87 | 609 | 7569 | 52983 |
| 90 - 94 | 2 | 92 | 184 | 8464 | 16928 |
| 95 - 99 | 1 | 97 | 97 | 9409 | 9409 |
| \(\sum\) | 40 | 3150 | 250590 |
(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
= \(\frac{3150}{40}\)
= 78.75
(b) \(SD = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\)
= \(\sqrt{\frac{250590}{40} - (\frac{3150}{40})^{2}}\)
= \(\sqrt{6264.75 - (78.75)^{2}}\)
= \(\sqrt{6264.75 - 6201.5625}\)
= \(\sqrt{63.1875}\)
= 7.95.