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 9.3 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.)

Answer 1

Answer:

a) 0.6062

b) 0.9505

c) 0.679

Step-by-step explanation:

The customer service center in a large new york department store has determined tha 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

(a) less than 10 minutes?

(b) longer than 5 minutes?

(c) between 8 and 15 minutes?

a) The Z score (z) is given by the equation:

$z=\frac{x-\mu}{\sigma}$,

Where:

μ is the mean = 9.3 minutes,

σ is the standard deviation = 2.6 minutes and x is the raw score

$z=\frac{x-\mu}{\sigma}=\frac{10-9.3}{2.6}=0.27$

From the z tables, P(X < 10) = P(z < 0.27) = 0.6062 = 60.62%

b) The Z score (z) is given by the equation:

$z=\frac{x-\mu}{\sigma}$,

$z=\frac{x-\mu}{\sigma}=\frac{5-9.3}{2.6}=-1.65$

From the z tables, P(X > 5) = P(z > -1.65) = 1 - P(z < -1.65) = 1 - 0.0495 =  0.9505 = 95.05%

c) For 8 minutes

$z=\frac{x-\mu}{\sigma}=\frac{8-9.3}{2.6}=-0.5$

For 15 minutes

$z=\frac{x-\mu}{\sigma}=\frac{15-9.3}{2.6}=2.19$

From the z tables, P(8< X < 15) = P(-0.5 < z < 2.19) = P(z < 2.19) - P(z< -0.5) = 0.9875 - 0.3085 = 0.679 = 67.9%