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

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

And we can use the z score formula given by:

`z = (\bar X -\mu)/(\sigma_(\bar X))`

And using the z score we have:

`P((3.5-4)/(0.3)< Z< (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<0.833) -P(Z<-1.667) = 0.798-0.048= 0.750`

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, (\sigma)/(√(n)))`

`\sigma_(\bar X)= (1.2)/(√(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 = (\bar X -\mu)/(\sigma_(\bar X))`

And using the z score we have:

`P((3.5-4)/(0.3)< Z< (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<0.833) -P(Z<-1.667) = 0.798-0.048= 0.750`

The graph illustrating the problem is on the figure attached.