Question

A major cell phone service provider has determined that the number of minutes per month that its customers use their phone is normally distributed with mean equal to 445.5 minutes and standard deviation equal to 22.75 minutes. The company is thinking of charging a lower rate for customers who use the phone the most. If it wishes to give the rate reduction to no more than 9 percent of its customers, what should the cut-off be?

Answer 1

we can say that the cut-off= 476.125 minutes

Explanation:

mean = 445.5 minutes

standard deviation = 22.75 minutes

rate reduction not more than = 9%

The z score corresponding to (100-9)%= 0.91 is 1.35

by standard normal Table

we use

`z=(x-\mu)/(\sigma)`

`1.35=(x-445.5)/(22.75)`

on calculating we get

x= 476.2125 minutes

Therefore, we can say that the cut-off= 476.125 minutes