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.
Verified Correct Answer: .
Step-by-Step Mathematical Solution
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) = \(\frac{(m + 5 + 3m - 4)}{23} = \frac{4m + 1}{23}\)
= \(\frac{4(3) + 1}{23} \)
= \(\frac{13}{23}\)
Source: Myschool Past Questions Archive (Record #52355)