Question

A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. the length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 3 minutes. find the waiting time at which only 10% of the customers will continue to hold.

Answer 1

The waiting time at which 10 percent of the people would continue to hold is given as 2.3

How to solve for the waiting time

We have to solve for X ~ Exponential(λ).

then E(X) = 1/λ = 3,

= 0.3333

Remember that the cumulative distribution function of X is F(x) = 1 - e^(-λx). ;  x is equal to the  time in over case

For 10 percent of the people we would have a probability of

10/100 = 0.1

we are to find

P(X ≤ t)

= 1 - e^(0.3333)(t) = 0.1

Our concern is the value of t

Then we take the like terms

1-0.1 = e^(0.3333)(t)

1/0.9 = e^(0.3333)(t)

t = 3 * ln(1/0.9)

= 0.3157