Question

A passenger arrives at a bus stop at 10am and waits for a bus that arrives at a time uniformly distributed between 10am and 10:30am. (a) What is the probability that the passenger waits more than 10 minutes? (b) What is the expected wait time? (c) What is the variance in wait time?

Answer #1

From provided information.

A passenger arrives at a bus stop at 10am and waits for a bus that arrives at a time uniformly distributed between 10am and 10:30am

Therefore,

X ~ Uniform ( a, b )

Where, a = 0 and b = 30

X ~ Uniform ( 0 , 30 )

(a) What is the probability that the passenger waits more than 10 minutes?

Answer :

P( X > 10 ) = 1- P( X ≤ 10)

P( X > 10 ) = 1 - 0.333 using uniform probability calculator.

P( X > 10 )= **0.667**

(b) What is the expected wait time?

Answer : E(X) = ( a + b) / 2

E(X) = ( 0+30)/2 = 15

Therefore, the expected wait time = 15 min

(c) What is the variance in wait time?

Answer : Var (X) = ( (b - a) ^2) / 12

Var (X) =((30-0)^2)/12

Var (X) = 900/ 12

Var (X) = 75 min 2

Therefore, the variance in wait time = 75 min2

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