Answer:
![P(3.25 < \bar X < 4.25)](https://tex.z-dn.net/?f=%20P%283.25%20%3C%20%5Cbar%20X%20%3C%204.25%29)
And we can use the z score formula given by:
![z = \frac{\bar X -\mu}{\sigma_{\bar X}}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20X%7D%7D)
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)](https://tex.z-dn.net/?f=%20P%28%5Cfrac%7B3.5-4%7D%7B0.3%7D%3C%20Z%3C%20%5Cfrac%7B4.25-4%7D%7B0.3%7D%29%20%3D%20P%28-1.667%3C%20Z%3C%200.833%29)
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](https://tex.z-dn.net/?f=P%28-1.667%3C%20Z%3C%200.833%29%3D%20P%28Z%3C0.833%29%20-P%28Z%3C-1.667%29%20%3D%200.798-0.048%3D%200.750)
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:
Where
and
And we select a sample size of n =16. Since the dsitribution for X is normal then the distribution for the sample mean
is also normal and given by:
![\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})](https://tex.z-dn.net/?f=%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%29)
![\sigma_{\bar X}= \frac{1.2}{\sqrt{16}}= 0.3](https://tex.z-dn.net/?f=%20%5Csigma_%7B%5Cbar%20X%7D%3D%20%5Cfrac%7B1.2%7D%7B%5Csqrt%7B16%7D%7D%3D%200.3)
For this case we want to find this probability:
![P(3.25 < \bar X < 4.25)](https://tex.z-dn.net/?f=%20P%283.25%20%3C%20%5Cbar%20X%20%3C%204.25%29)
And we can use the z score formula given by:
![z = \frac{\bar X -\mu}{\sigma_{\bar X}}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Cbar%20X%20-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20X%7D%7D)
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)](https://tex.z-dn.net/?f=%20P%28%5Cfrac%7B3.5-4%7D%7B0.3%7D%3C%20Z%3C%20%5Cfrac%7B4.25-4%7D%7B0.3%7D%29%20%3D%20P%28-1.667%3C%20Z%3C%200.833%29)
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](https://tex.z-dn.net/?f=P%28-1.667%3C%20Z%3C%200.833%29%3D%20P%28Z%3C0.833%29%20-P%28Z%3C-1.667%29%20%3D%200.798-0.048%3D%200.750)
The graph illustrating the problem is on the figure attached.