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Question 26 University of Oshawa Tech has conducted a survey of its faculty members and has found the following distribution

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Answer #1

a) P(Four or more) = 1 - (0.45 + 0.25 + 0.15) = 0.15

b) Average cost = E(X) = 0.45*130 + 0.25*200 + 0.15*270 + 0.15*350 = $ 201.50

c) E(X​​​​​​2) = (0.45)*1302 + 0.25*2002 + 0.15*2702 + 0.15*3502 = 46915

Hence,

Variance = E(X​​​​​​2) - [E(X)]2 = 46915 - 210.502 = 6312.75

So,

Standard deviation = 6312.75 = $ 79.45

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