Theory question 110
Measures of DispersionThe table shows the distribution of marks scored by students in an examination.
Class Interval: 60 − 64, 65 − 69, 70 − 74, 75 − 79, 80 − 84, 85 − 89, 90 − 94, 95 − 99
Frequency: 2, 3, 6, 11, 8, 7, 2, 1
Calculate, correct to 2 decimal places, the (a) mean; (b) standard deviation of the distribution.
Show the model answer
Model answer
Class marks: 62, 67, 72, 77, 82, 87, 92, 97; Σf = 40.
(a) Σfx = 124 + 201 + 432 + 847 + 656 + 609 + 184 + 97 = 3150. Mean = 3150/40 = 78.75.
(b) Σf(x − 78.75)2 = 561.13 + 414.19 + 273.38 + 33.69 + 84.50 + 476.44 + 351.13 + 333.06 = 2527.5.
SD = √(2527.5/40) = √63.19 ≈ 7.95.
Explanation
Use the class mid-points as x.