Question

A wireless communication company is considering switching from a per-minute charge to a flat monthly fee for unlimited service. It anticipates that this will result in greater usage and would like to estimate what the average number of minutes per month per customer will be under the new plan. To do this, it offers 1000 customers the flat-fee plan for one month and tracks their usage. From past data, the company knows that the standard deviation of monthly minutes is about 120 minutes; the company expects that this figure will not change much under the new plan. Find the probability that the average among the 1000 test customers will differ from the true mean by more than 5 minutes.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 0

standard deviation = $\sigma$ = 120

n = 1000

$\mu$_{$\bar x$} =  $\mu$ = 0

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 120 / $\sqrt$ 1000 = 3.79

P(-5 < < 5)  

T

= 1 - P[(-5 - 0) / 3.79 < (
T
- $\mu$ _{$\bar x$}) / $\sigma$ _{$\bar x$} < (5 - 0) / 3.79)]

= 1 - P( -1.32 < Z < 1.32 )

= 1 - P(Z < 1.32) - P(Z < -1.32)

Using z table,  

= 1 - P( 0.9066 - 0.0934)   

= 1 - 0.8132

= 0.1868