104k views
1 vote
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 =________.

1 Answer

4 votes

Answer:

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.

User Francesco Dondi
by
8.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories