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 time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. what is the probability that a randomly selected caller is placed on hold fewer than 7 minutes?

0.917915

Step-by-step explanation:

Exponential distribution is expressed as it follows

F (y,λ) = 1 – e ^(- λ(y) )

Where λ is equal to 1/ population mean

So λ = 1/2.8 = 0.357

Let y be the length of time a costumer wait on hold

Now, let´s find the value for y = 7 minutes, which is the proportion of costumer that wait on hold

P (y<7) = 1 – e^(- .357*(7) )

P (y<7) = 0.917915

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

594 square inches

step-by-step explanation:

vote !