| Marks | 1 | 2 | 3 | 4 | 5 |
| Number of students | m + 2 | m - 1 | 2m - 3 | m + 5 | 3m - 4 |
The table shows the distribution of marks scored by some students in a test.
(a) If the mean mark is \(3\frac{6}{23}\), find the value of m.
(b) Find the : (i) interquartile range
(ii) probability of selecting a student who scored at least 4 marks in the test.
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Correct Answer: Option n
Explanation:
(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
\(\frac{75}{23} = \frac{28m - 9}{8m - 1}\)
\(75(8m - 1) = 23(28m - 9) \implies 600m - 75 = 644m - 207\)
\(-75 + 207 = 644m - 600m\)
\(132 = 44m \implies m = 3\)
(b)(i) Interquartile range = Third quartile - First quartile
Frequency = 8(3) - 1 = 24 - 1 = 23
\(Q_{3} = \frac{3}{4} \times 23 = 17.25th\) position = 4
\(Q_{1} = \frac{1}{4} \times 23 = 5.75th\) position = 2
Interquartile range : 4 - 2 = 2
(ii) P(at least 4 marks) = \(\frac{(m + 5 + 3m - 4)}{23} = \frac{4m + 1}{23}\)
= \(\frac{4(3) + 1}{23} \)
= \(\frac{13}{23}\)
| Mark (x) | 1 | 2 | 3 | 4 | 5 | Total |
| Frequency (f) | m + 2 | m - 1 | 2m - 3 | m + 5 | 3m - 4 | 8m - 1 |
| fx | m + 2 | 2m - 2 | 6m - 9 | 4m + 20 | 15m - 20 | 28m - 9 |
(a) Mean \(\bar{x} = \frac{\sum fx}{\sum f}\)
\(\frac{75}{23} = \frac{28m - 9}{8m - 1}\)
\(75(8m - 1) = 23(28m - 9) \implies 600m - 75 = 644m - 207\)
\(-75 + 207 = 644m - 600m\)
\(132 = 44m \implies m = 3\)
(b)(i) Interquartile range = Third quartile - First quartile
Frequency = 8(3) - 1 = 24 - 1 = 23
\(Q_{3} = \frac{3}{4} \times 23 = 17.25th\) position = 4
\(Q_{1} = \frac{1}{4} \times 23 = 5.75th\) position = 2
Interquartile range : 4 - 2 = 2
(ii) P(at least 4 marks) = \(\frac{(m + 5 + 3m - 4)}{23} = \frac{4m + 1}{23}\)
= \(\frac{4(3) + 1}{23} \)
= \(\frac{13}{23}\)