Question

the manager of a supermarket tracked the amount of time needed for customers to be served by the cashier. the manager concluded that the checkout times are exponentially distributed with a mean of 4 minutes. what proportion of customers require more than 10 minutes to check out?

Answer 1

Explanation:

the probability density function is then

f(x, lambda) = lambda × e^(- lambda×x)

lambda being the rate parameter and is equal to 1/mean.

as in our case the mean value is 4 we have

f(x, ¼) = ¼×e^(-¼x) = ¼×e^(-x/4)

the integral between 0 and n over the probability density function gives us the cumulative density function and with that the probability that x < n.

this integral function (cumulative density function) is

F(x) = 1 - e^(- lambda×x)

in our case

F(x) = 1 - e^(-x/4)

the probability (= %/100) of a customer to take longer than 10 minutes is

P(x > 10) = 1 - P(x < 10) = 1 - (1 - e^(-10/4)) = e^(-2.5) =

= 0.082084999...

the proportion of customers taking longer than 10 minutes is then

P(x > 10)×100 % = 0.082084999...×100 % =

= 8.2084999...% ≈ 8.21%