The table gives the frequency distribution of marks obtained by a group of students in a test.
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 |
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.
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Correct Answer: Option n
Explanation:
\(\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\)
\(\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\)
\(\bar{x} = \frac{\sum fx}{\sum f}\)
\(5 = \frac{101 + 9x}{18 + 2x} \implies 101 + 9x = 5(18 + 2x)\)
\(101 + 9x = 90 + 10x \implies 101 - 90 = 10x - 9x\)
\(11 = x \)
(b)(i) Mode = 5.
(ii) Median
Frequency = 18 + 2(11) = 40.
Median position = \(\frac{40}{2} = 20\)
20th position = 5.
(c) No of students that scored at least 7 marks = 4 + 1 = 5.
Probability of scoring at least 7 marks = \(\frac{5}{40} = \frac{1}{8}\)
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | x - 1 | x | 9 | 4 | 1 |
| fx | 15 | 4x - 4 | 5x | 54 | 28 | 8 |
\(\sum fx = 15 + 4x - 4 + 5x + 54 + 28 + 8 = 101 + 9x\)
\(\sum f = 5 + x - 1 + x + 9 + 4 + 1 = 18 + 2x\)
\(\bar{x} = \frac{\sum fx}{\sum f}\)
\(5 = \frac{101 + 9x}{18 + 2x} \implies 101 + 9x = 5(18 + 2x)\)
\(101 + 9x = 90 + 10x \implies 101 - 90 = 10x - 9x\)
\(11 = x \)
(b)(i) Mode = 5.
(ii) Median
Frequency = 18 + 2(11) = 40.
Median position = \(\frac{40}{2} = 20\)
20th position = 5.
(c) No of students that scored at least 7 marks = 4 + 1 = 5.
Probability of scoring at least 7 marks = \(\frac{5}{40} = \frac{1}{8}\)