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THE UNIVERSITY OF HONG KONG | DEPARTMENT OF STATISTICS AND ACTURIAL SCIENCE | STAT 0302 Business Statistics | April 2005 Class Test solutions

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTURIAL SCIENCE STAT 0302 Business Statistics April 2005 Class Test solutions 1. Class 8-12 13-17 18-22 23-27 28-32 33-37 Class boundar ... ies 7.5-12.5 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5 x Class mark 10 15 20 25 30 35 Frequency 2 5 7 13 8 4 Cumulative frequency 2 7 14 27 35 39 Mean = N ∑f i x i = (10x2) + (15x5) + (20x7) + (25x13) + (30x8) + (35x4) µ = 24.1025641 S.D. = σ = 2 2 2 24.1025641 39 24350 − = − ∑ µ N xi =6.589793475 3 3 1 1 3 3 2 f f N Median L c i ∑ i − = − = + where L3 = left boundary of the median class = 22.5 3 c = length of the median class = 5 3 f = frequency in the median class = 13 ∑ − = 3 1 i 1 i f = total frequency before the median = 14 N = number of data = 39. Median = 22.5 + (27.5 – 22.5)x 13 14 2 39.5 − = 24.61538462 13 5.5 22.5 + 5x = Skewness = 6.589793475 3( ) 3(24.1025641− 24.61538462) = − σ µ median = -23.35% (negatively skewed) 1 ANSWER: E 2. Data set 1 Data set 2 N1 = 10 N2 = 17 µ1 = 29 µ2 = 31 6 2 σ 1 = σ 2 = 2.5 Enter the data into the calculator as: 29 − 6 29 + 6 31-2.5=28.5 31+2.5=33.5 5 5 17/2=8.5 8.5 [Show More]

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