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 4 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 be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

Answer 1

Answer:

$P(3.25 < \bar X < 4.25)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

And using the z score we have:

$P(\frac{3.5-4}{0.3}< Z< \frac{4.25-4}{0.3}) = P(-1.667< Z< 0.833)$

And we can use the normal standard distribution table or excel in order to find the probabilities and we got:

$P(-1.667< Z< 0.833)= P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the time it takes her to complete one review, and for this case we know the distribution for X is given by:

$X \sim N(4,1.2)$  

Where $\mu=4$ and $\sigma=1.2$

And we select a sample size of n =16. Since the dsitribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\sigma_{\bar X}= \frac{1.2}{\sqrt{16}}= 0.3$

For this case we want to find this probability:

$P(3.25 < \bar X < 4.25)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

And using the z score we have:

$P(\frac{3.5-4}{0.3}< Z< \frac{4.25-4}{0.3}) = P(-1.667< Z< 0.833)$

And we can use the normal standard distribution table or excel in order to find the probabilities and we got:

$P(-1.667< Z< 0.833)= P(Z$

The graph illustrating the problem is on the figure attached.