Theory question 106
Measures of Central TendencyThe 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 mark 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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(a) Σf = 2x + 18; Σfx = 15 + 4(x − 1) + 5x + 54 + 28 + 8 = 9x + 101.
(9x + 101)/(2x + 18) = 5 gives 9x + 101 = 10x + 90, so x = 11.
(b) Frequencies are 5, 10, 11, 9, 4, 1 (total 40).
(i) Mode = 5 (highest frequency, 11).
(ii) The 20th and 21st marks are both 5 (cumulative 5, 15, 26…), so median = 5.
(c) At least 7 marks: 4 + 1 = 5 students. P = 5/40 = 1/8.
Set up mean = Σfx ÷ Σf with x still unknown, then solve.