The table shows the marks scored by a group of students in a class test.
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 1 | 4 | 9 | 8 | 5 | 3 |
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
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Correct Answer: Option n
Explanation:
(a)
(i) Mean(\(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{81}{30}\)
= 2.7
(ii) \(N = \sum f = 30\)
\(median = \frac{1}{2}(N + 1)\)
= \(\frac{1}{2}(30 + 1)\)
=15.5
= The 15th and 16th position.
\(\therefore median = \frac{3 + 3}{2} = 3\)
(b) Sectorial angle for the mark 2 = \(\frac{9}{30} \times 360°\)
= 108°.
(a)
| x | 0 | 1 | 2 | 3 | 4 | 5 | Total |
| f | 1 | 4 | 9 | 8 | 5 | 3 | 30 |
| fx | 0 | 4 | 18 | 24 | 20 | 15 | 81 |
(i) Mean(\(\bar{x}\) = \(\frac{\sum fx}{\sum f}\)
= \(\frac{81}{30}\)
= 2.7
(ii) \(N = \sum f = 30\)
\(median = \frac{1}{2}(N + 1)\)
= \(\frac{1}{2}(30 + 1)\)
=15.5
= The 15th and 16th position.
\(\therefore median = \frac{3 + 3}{2} = 3\)
(b) Sectorial angle for the mark 2 = \(\frac{9}{30} \times 360°\)
= 108°.