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Using the assumed-mean method with A = 150, find the mean of: classes 100–120, 120–140, 140–160, 160–180, 180–200 with frequencies 6, 10, 14, 12, 8.

Competition practiceStatistics

A150
B147.6
C152
D152.4

Step-by-Step Solution

d = x - 150 : - 40, - 20, 0, 20, 40. \sum fd = - 240 - 200 + 0 + 240 + 320 = 120, N = 50

; mean

= 150 + \frac{120}{50} = 152.4

.

Answer: D) 152.4

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