The complete question is;
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 8.9 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:
A) P (x < 10) = 0.6700
B) P (x > 5 ) = 0.9406
C) P (8.0000 < x < 15.0000) = 0.6332
Explanation:
A) we are given;
Mean;μ = 8.9 minutes
Standard deviation;σ = 2.5 minutes
Normal random variable;x = 10
So to find;P(x < 10) we will use the Z-score formula;
z = (x - μ)/σ
z = (10 - 8.9)/2.5 = 0.44
From z-distribution table and Z-score calculator as attached, we have;
P (x < 10) = P (z < 0.44) = 0.6700
B) similarly;
z = (x - μ)/σ =
z = (5 - 8.9)/2.5
z = -1.56
From z-distribution table and Z-score calculator as attached, we have;
P (x > 5 ) = P (z > -1.56) = 0.9406
C)between 8 and 15 minutes
For 8 minutes;
z = (8 - 8.9)/2.5 = -0.36
For 15 minutes;
z = (15 - 8.9)/2.5 = 2.44
From z-distribution table and Z-score calculator as attached, we have;
P (8.0000 < x < 15.0000) = P (-0.36 < z < 2.44) = 0.6332