| 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 (3frac{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.
Explanation
| 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})