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 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 ). What proportion of customers having to hold more than 1.5 minutes will hang up before placing an order? A) 0.86466 B) 0.60653 C) 0.39347 D) 0.13534

Answer 1

Answer:

B) 0.60653

Step-by-step explanation:

The exponential distribution has one parameter which is lambda denoted as λ.

λ = 1/β; where β represents the population mean.

The cumulative distribution function is:

F(X;λ) = $1 - e^{-lambdax}$

$=1 - e^{-\frac{1}{\beta}x}$

The average waiting time is β = 3 min.

The proportion of customers willing to wait on hold for more than 1.5 minutes is:

P(X>1.5) = 1-P(X$\leq$1.5)

$= 1-(1-e^{-1.5*(1/3)} )\\= 1-1+e^{-1.5/3} \\=0.60653$