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.)

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

A) P (x < 10) = 0.6700

B) P (x > 5 ) = 0.9406

C) P (8.0000 < x < 15.0000) = 0.6332

Step-by-step 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

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

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