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 8.3 minutes and a standard deviation of 2.0 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 = = 8.3

standard deviation = =2.0

(a)

P(x < 10) = P[(x - ) / < (10 - 8.3) / 2.0]

= P(z < 0.85)

= 0.8023

probability = 0.8023

(b)

P(x > 5) = 1 - P(x < 5)

= 1 - P((x - ) / < (5 - 8.3) / 2.0)

= 1 - P(z < -1.65)

= 1 - 0.0495

= 0.9505   

  Probability = 0.9505

(c)

P(8 < x < 15) = P((8 - 8.3)/ 2.0) < (x - ) / < (15 - 8.3) / 2.0) )

= P(-0.15 < z < 3.35)

= P(z < 3.35) - P(z < -0.15)

= 0.9996 - 0.4404

= 0.5592

Probability = 0.5592