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

Answer:

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

b) The 95th percentile is 4.4935 hours.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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 $n = 16, s = \frac{1.2}{\sqrt{16}} = 0.3$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{4.25 - 4}{0.3}$

$Z = 0.83$

$Z = 0.83$ has a pvalue of 0.7967

X = 3.5

$Z = \frac{X - \mu}{s}$

$Z = \frac{3.5 - 4}{0.3}$

$Z = -1.67$

$Z = -1.67$ 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.

$Z = \frac{X - \mu}{s}$

$1.645 = \frac{X - 4}{0.3}$

$X - 4 = 0.3*1.645$

$X = 4.4935$

The 95th percentile is 4.4935 hours.