The frequency table shows the marks scored by 32 students in a test. Marks scored 1 2 3 4 5 6 7 8 9 10 No of students 2 3 4 4 4 4 5 3 2 1 Find the : (a)(i) ...
Question 1 Report
The frequency table shows the marks scored by 32 students in a test.
| Marks scored | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
Find the :
(a)(i) mean ; (ii) median ; (iii) mode of the marks;
(b) percentage of the students who scored at least 8 marks.
Answer Details
Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).
| Mark (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
| fx | 2 | 6 | 12 | 16 | 20 | 24 | 35 | 24 | 18 | 10 |
| Cum. f | 2 | 5 | 9 | 13 | 17 | 21 | 26 | 29 | 31 | 32 |
(a)(i) Mean. \(\sum fx = 167\).
\[\bar{x}=\frac{167}{32}\approx 5.22.\]
(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.
\[\text{Median}=\frac{5+5}{2}=5.\]
(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.
(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.
\[\frac{6}{32}\times100\% = 18.75\%.\]
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