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 ...
Question 1 Report
| 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.
Answer Details
(a) Finding m. Form the totals in terms of m.
\[\sum f=(m+2)+(m-1)+(2m-3)+(m+5)+(3m-4)=8m-1.\]
\[\sum fx=1(m+2)+2(m-1)+3(2m-3)+4(m+5)+5(3m-4)=28m-9.\]
The mean is \(3\frac{6}{23}=\frac{75}{23}\), so
\[\frac{28m-9}{8m-1}=\frac{75}{23}.\]
\[23(28m-9)=75(8m-1)\Rightarrow 644m-207=600m-75\Rightarrow 44m=132\Rightarrow m=3.\]
Substituting \(m=3\) gives the frequencies:
| Mark | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Frequency | 5 | 2 | 3 | 8 | 5 |
| Cumulative | 5 | 7 | 10 | 18 | 23 |
(b)(i) Interquartile range. \(N=23\). \(Q_1\) is the \(\frac{N+1}{4}=6\text{th}\) value \(=2\); \(Q_3\) is the \(\frac{3(N+1)}{4}=18\text{th}\) value \(=4\).
\[\text{IQR}=Q_3-Q_1=4-2=2.\]
(b)(ii) Probability of at least 4 marks.
\[P(\ge4)=\frac{8+5}{23}=\frac{13}{23}.\]
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