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. Let X be the random variable representing the time it takes her to complete one review. Assume X is normally distributed. Let X be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews. Complete the distributions.

A. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs.
B. P(_____) = _______.
C. Find the 95th percentile for the mean time to complete one month's reviews.
D.The 95th Percentile =________.

Answer 1

a)
`P(3.5 \leq X \leq 4.25) = 0.7492`

b) The 95th percentile is 4.4935 hours.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours.

This means that
`\mu = 4, \sigma = 1.2`

16 reviews.

This means that
`n = 16, s = (1.2)/(√(16)) = 0.3`

A. Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs.

This is the pvalue of Z when X = 4.25 subtracted by the pvalue of Z when X = 3.5. So

X = 4.25

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (4.25 - 4)/(0.3)`

`Z = 0.83`

`Z = 0.83` has a pvalue of 0.7967

X = 3.5

`Z = (X - \mu)/(s)`

`Z = (3.5 - 4)/(0.3)`

`Z = -1.67`

`Z = -1.67` has a pvalue of 0.0475

0.7967 - 0.0475 = 0.7492

So

`P(3.5 \leq X \leq 4.25) = 0.7492`

C. Find the 95th percentile for the mean time to complete one month's reviews.

This is X when Z has a pvalue of 0.95, so X when Z = 1.645.

`Z = (X - \mu)/(s)`

`1.645 = (X - 4)/(0.3)`

`X - 4 = 0.3*1.645`

`X = 4.4935`

The 95th percentile is 4.4935 hours.