Acatalog 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

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) =

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(X1.5)

typically if you want , you would actually put questions here for us to you. but hey, that's just me. maybe someone will magically answer the questions for you without knowing what they are.

45

step-by-step explanation:

to find a percentage, add a decimal in front and multiply it by the other number.

.25 times 180= 45