Question

Suppose your waiting time for a bus in the morning is uniformly distributed on [0, 8], whereas waiting time in the evening is

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Answer #1

since for uniform distribution: U(a,b)

mean =(a+b)/2 and Variance =(b-a)2/12

a) total expected waiting time for a week =5*((0+8)/2+(0+10)/2) =45

b)

Variance =5*((8-0)^2/12+(10-0)^2/12)=68.33

c)

expected value=(0+8)/2-(0+10)/2 =-1

Variance =((8-0)^2/12+(10-0)^2/12)=13.67

d)

expected value = -5

Variance =68.33

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