The table gives the frequency distribution of marks obtained by a group of students in a test.
| Marks |
3 |
4 |
5 |
6 |
7 |
8 |
| Frequency |
5 |
\(x - 1\) |
\(x\) |
9 |
4 |
1 |
If the mean is 5,
(a) Calculate the value of x;
(b) Find the : (i) mode ; (ii) median of the distribution.
(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | x-1 | x | 9 | 4 | 1 |
|---|
(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):
\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]
Since mean \(=5\):
\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]
So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).
(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.
(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.
(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]