Question

2. 8 points BBBasicStat6 7.R.016. My Notes 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 9.3 minutes and a standard deviation of 2.5 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 = $\mu$ = 9.3

standard deviation = $\sigma$ = 2.5

a) P(x < 10 ) = P[(x - $\mu$ ) / $\sigma$ < (10 - 9.3) /2.5 ]

= P(z < 0.28 )

= 0.6103

probability =0.6103

b)

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

=1- p [(x - $\mu$ ) / $\sigma$ < (5 - 9.3) /2.5 ]

=1- P(z < -1.72)

= 1 - 0.0427 = 0.9573

probability = 0.9573

c)

P( 8< x < 15 ) = P[(8 -9.3)/2.5 ) < (x - $\mu$ ) /$\sigma$  < (15 - 9.3) /2.5 ) ]

= P( -0.52< z < 2.28 )

= P(z < 2.28 ) - P(z < -0.52 )

= 0.9887 - 0.3015 = 0.6872

Probability = 0.6872