Question

During lunch hour, customers arrive at a fast food drive-through window, on average, every

2.9

minutes. Assume arrival time follows the exponential probability distribution.

a) What is the probability that the next customer will arrive within the next

2

minutes?

b) What is the probability that the next customer will arrive in more than

7

minutes?

c) What is the probability that the next customer will arrive between

1

and

6

minutes?

Answer 1

X ~ Exponential ()

Where = 2.9

a)

P(X < x) = 1 - e^{-X /}

So,

P( X < 2) = 1 - e^{-2 / 2.9}

= 0.4983

b)

P( X > x) = e^{-X /}

So,

P( X > 7) = e^{-7 / 2.9}

= 0.0895

c)

P( x1 < X < x2) = e^{-X1 /} - e^{-X2 /}

So,

P( 1 < X < 6) = e^{-1 / 2.9} - e^{- 6 / 2.9}

= 0.7083 - 0.1263

= 0.5820