Question

The 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 9.1 minutes and a standard deviation of 2.2 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

= 9.1, = 2.2

(a) less than 10 minutes

we want to find P(x < 10)

z= 0.4091

P(x < 10) = P(z < 0.4091 )

now find   P(z < 0.4091 ) using normal z table

P(z < 0.4091 ) =  0.6588

P(x < 10) = 0.6588

less than 10 minutes =  0.6588

(b) longer than 5 minutes

we want to find P(x > 5)

z= −1.8636

P(x >5) = 1 - P(z < −1.8636 )

now find   P(z < −1.8636 ) using normal z table

P(z < −1.8636 ) =  0.0312

P(x >5) = 1 -  0.0312

P(x >5) = 0.9688

longer than 5 minutes = 0.9688

(c) between 8 and 15 minutes

we want to find P(8 < x < 15)

P(8 < x < 15) = P(x < 15) - P(x < 8)

first find P(x < 15)

z= 2.6818

P(x < 15) = P(z < 2.6818 )

now find   P(z < 2.6818 ) using normal z table

P(z < 2.6818 ) =  0.9963

P(x < 15) = 0.9963

first find P(x < 8)

z= −0.5

P(x < 15) = P(z < −0.5 )

now find   P(z < −0.5 ) using normal z table

P(z < −0.5 ) =  0.3085

P(x < 8) = 0.3085

P(8 < x < 15) = P(x < 15) - P(x < 8)

P(8 < x < 15) = 0.9963−0.3085

P(8 < x < 15) =0.6878

between 8 and 15 minutes = 0.6878