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 me...
Question 1 Report
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.
Answer Details
| 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 \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]Download The App On Google Playstore
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