Answer:
[tex] P(3.25 < \bar X < 4.25)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]
And using the z score we have:
[tex] P(\frac{3.5-4}{0.3}< Z< \frac{4.25-4}{0.3}) = P(-1.667< Z< 0.833)[/tex]
And we can use the normal standard distribution table or excel in order to find the probabilities and we got:
[tex]P(-1.667< Z< 0.833)= P(Z<0.833) -P(Z<-1.667) = 0.798-0.048= 0.750[/tex]
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:
[tex]X \sim N(4,1.2)[/tex]
Where [tex]\mu=4[/tex] and [tex]\sigma=1.2[/tex]
And we select a sample size of n =16. Since the dsitribution for X is normal then the distribution for the sample mean [tex] \bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \sigma_{\bar X}= \frac{1.2}{\sqrt{16}}= 0.3[/tex]
For this case we want to find this probability:
[tex] P(3.25 < \bar X < 4.25)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\sigma_{\bar X}}[/tex]
And using the z score we have:
[tex] P(\frac{3.5-4}{0.3}< Z< \frac{4.25-4}{0.3}) = P(-1.667< Z< 0.833)[/tex]
And we can use the normal standard distribution table or excel in order to find the probabilities and we got:
[tex]P(-1.667< Z< 0.833)= P(Z<0.833) -P(Z<-1.667) = 0.798-0.048= 0.750[/tex]
The graph illustrating the problem is on the figure attached.
