Theory question 107
Measures of DispersionThe weights to the nearest kilogram of a group of 50 students in a College of Technology are given below:
65, 70, 60, 46, 51, 55, 59, 63, 68, 53
47, 53, 72, 53, 67, 62, 64, 70, 57, 56
73, 56, 48, 51, 58, 63, 65, 62, 49, 64
53, 59, 63, 50, 48, 72, 67, 56, 61, 64
66, 52, 49, 62, 71, 58, 53, 69, 63, 59
(a) Prepare a grouped frequency table with class intervals 45 − 49, 50 − 54, 55 − 59, etc.
(b) Using an assumed mean of 62 or otherwise, calculate the mean and the standard deviation of the grouped data, correct to one decimal place.
Show the model answer
(a) 45−49: 6; 50−54: 9; 55−59: 10; 60−64: 12; 65−69: 7; 70−74: 6 (total 50)
(b) Class marks 47, 52, 57, 62, 67, 72; d = x − 62: −15, −10, −5, 0, 5, 10
Σfd = −90 − 90 − 50 + 0 + 35 + 60 = −135; mean = 62 − 135/50 = 59.3kg
Σfd2 = 1350 + 900 + 250 + 0 + 175 + 600 = 3275
SD = √(3275/50 − (−2.7)2) = √(65.5 − 7.29) = √58.21 = 7.6kg
Use tally marks so that no weight is missed or counted twice.