Question

Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours.

(1)  What is the mean time to complete one review?

(2)  What is the standard deviation to complete one review?

(3) What is the probability that one review will take Yoonie from 3.5 to 4.25 hours? (Round to 3 decimal places)

(4) What is the mean for the average time for Yoonie to complete the 16 reviews?

(5) What is the standard deviation for her average time to complete the 16 reviews?

(6) What is the probability that 16 reviews will take Yoonie from 3.5 to 4.25 hours? (Round to three decimal places)

Answer 1

a)

mean = 4

2)

std.deviation = 1.2

3)

Here, μ = 4, σ = 1.2, x1 = 3.5 and x2 = 4.25. We need to compute P(3.5<= X <= 4.25). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (3.5 - 4)/1.2 = -0.42
z2 = (4.25 - 4)/1.2 = 0.21

Therefore, we get
P(3.5 <= X <= 4.25) = P((4.25 - 4)/1.2) <= z <= (4.25 - 4)/1.2)
= P(-0.42 <= z <= 0.21) = P(z <= 0.21) - P(z <= -0.42)
= 0.5832 - 0.3372
= 0.246

4)

mean = 4

5)

std. deviation = 1.2/sqrt(16) = 0.3

6)
Here, μ = 4, σ = 0.3, x1 = 3.5 and x2 = 4.25. We need to compute P(3.5<= X <= 4.25). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (3.5 - 4)/0.3 = -1.67
z2 = (4.25 - 4)/0.3 = 0.83

Therefore, we get
P(3.5 <= X <= 4.25) = P((4.25 - 4)/0.3) <= z <= (4.25 - 4)/0.3)
= P(-1.67 <= z <= 0.83) = P(z <= 0.83) - P(z <= -1.67)
= 0.7967 - 0.0475
= 0.749