Question

A 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 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 (i.e. the mean number of calls answered in a minute is 1/3). What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? (Hints: Expoential distribution) A. 0.48658 B. 0.22313 C. 0.51342 D. 0.77687

Answer 1

Explanation:

`m = (1)/(3)`

Probability of more than time more than 4.5 minutes

=
`e^(-mx)`

`m = (1)/(3)`

x = 4.5

probability required.

=
`e^{-(1)/(3)* 4.5}`

=
`e^(-1.5)`

= .22313