Theory question 113
Measures of DispersionThe following are the marks scored by 20 students in a test:
15, 11, 17, 25, 13, 15, 16, 22, 24, 27,
20, 22, 15, 16, 15, 19, 22, 24, 22, 11
(a) Prepare a frequency table for the distribution using class intervals 10 − 12, 13 − 15, 16 − 18 … .
(b) Calculate the variance of the distribution.
(c) If the pass mark for the test was 16, find the probability, that a student selected at random from the class failed.
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(a) 10–12: 2; 13–15: 5; 16–18: 3; 19–21: 2; 22–24: 6; 25–27: 2 (total 20)
(b) Class marks 11, 14, 17, 20, 23, 26; Σfx = 373, mean = 18.65
Σf(x − x̄)² = 117.05 + 108.11 + 8.17 + 3.65 + 113.54 + 108.05 ≈ 458.55; variance = 458.55 ÷ 20 ≈ 22.93
(c) Students scoring below 16: 11, 11, 13, 15, 15, 15, 15 = 7, so P(failed) = 7/20
Use the class marks (mid-points) as x.