Question

Assume the hold time of callers to a cable company is normally distributed with a mean of 20 minutes and a standard deviation of 0.5 minute. Determine the percent of callers who are on hold for more than 20 minutes Click here to view page 1 of the standard normal distrbution table, Click here to view page 2 of the standard romal distribution table, % (Round to two decimal places as needed)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 2.0

standard deviation = $\sigma$ = 0.5

P(x > 2.0 ) = 1 - P(x < 2.0)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (2.0-2.0) /0.5 ]

= 1 - P(z < 0)

=0.50

50.00 % callers hold for more than 2.0 minutes