Question

he Customer Service Center in a large New York department store has determined that the amount of time spent with a customer about a complaint is normally distributed, with a mean of 8.5 minutes and a standard deviation of 2.6 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint will be as follows. (Round your answers to four decimal places.) (a) less than 10 minutes (b) longer than 5 minutes (c) between 8 and 15 minutes

Answer 1

Solution :

Given that ,

mean = = 8.5

standard deviation = =2.6

P(X< 10) = P[(X- ) / < (10-8.5) /2.6 ]

= P(z <0.58 )

Using z table

= 0.7190

B.

P(X> 5) = 1 - P[(X- ) / < (5-8.5) /2.6 ]

=1- P(z < -1.35)

Using z table

= 1 - 0.0885

=0.9115   

C.

P(8< x < 115) = P[(8-8.5) / 2.6< (x - ) / < (10-8.5) /2.6 )]

= P(-0.19 < Z <0.58 )

= P(Z < 0.58) - P(Z <-0.19 )

Using z table   

= 0.7190-0.4247

probability= 0.2943