Question

a company that receives the majority of its orders by telephone conducted a study to determing how long customers were willing to wait on hold before ordering a product. The length of waiting time was found to be a variable best approximated by an exponential distribution with a mean length of waiting time equal to 3 minutes. What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order 0.51342 0.48658

Answer 1

0.2231 = 22.31% of customers having to hold more than 4.5 minutes will hang up before placing an order

Explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

`f(x) = \mu e^(-\mu x)`

In which
`\mu = (1)/(m)` is the decay parameter.

The probability that x is lower or equal to a is given by:

`P(X \leq x) = \int\limits^a_0 {f(x)} \, dx`

Which has the following solution:

`P(X \leq x) = 1 - e^(-\mu x)`

The probability of finding a value higher than x is:

`P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^(-\mu x)) = e^(-\mu x)`

The length of waiting time was found to be a variable best approximated by an exponential distribution with a mean length of waiting time equal to 3 minutes.

This means that
`m = 3, \mu = (1)/(3)`

What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order?

This is:

`P(X > 4.5) = e^{-(1)/(3)*4.5} = e^{-(4.5)/(3)} = e^(-1.5) = 0.2231`

0.2231 = 22.31% of customers having to hold more than 4.5 minutes will hang up before placing an order