Answer:
a) ![P(X](https://tex.z-dn.net/?f=P%28X%3C67.5%29%20%3DP%28Z%3C%20%5Cfrac%7B67.5-65%7D%7B5%7D%29%20%3D%20P%28Z%3C0.5%29)
And we can find this probability using the z table or excel:
![P(z](https://tex.z-dn.net/?f=P%28z%3C0.5%29%3D0.691)
b) ![P(64.3](https://tex.z-dn.net/?f=P%2864.3%3C%5Cbar%20X%3C66.4%29%3DP%28%5Cfrac%7B64.3-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3C%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3C%5Cfrac%7B64.3-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%29%3DP%28%5Cfrac%7B64.3-65%7D%7B1.066%7D%3CZ%3C%5Cfrac%7B66.4-65%7D%7B1.066%7D%29%3DP%28-0.657%3Cz%3C1.313%29)
![P(-0.657](https://tex.z-dn.net/?f=P%28-0.657%3Cz%3C1.313%29%3DP%28z%3C1.313%29-P%28z%3C-0.657%29)
![P(-0.657](https://tex.z-dn.net/?f=P%28-0.657%3Cz%3C1.313%29%3DP%28z%3C1.313%29-P%28z%3C-0.657%29%3D0.905-0.256%3D0.650)
c) And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.
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".
Part a
Let X the random variable that represent the interpupillary distance of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=5](https://tex.z-dn.net/?f=%5Csigma%3D5)
We are interested on this probability
![P(X](https://tex.z-dn.net/?f=P%28X%3C67.5%29)
And the best way to solve this problem is using the normal standard distribution and the z score given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we apply this formula to our probability we got this:
![P(X](https://tex.z-dn.net/?f=P%28X%3C67.5%29%20%3DP%28Z%3C%20%5Cfrac%7B67.5-65%7D%7B5%7D%29%20%3D%20P%28Z%3C0.5%29)
And we can find this probability using the z table or excel:
![P(z](https://tex.z-dn.net/?f=P%28z%3C0.5%29%3D0.691)
Part b
For this case the distirbution of the sample mean is also normal since the distribution for the random variable X is normal and is given by:
![\bar X \sim N(\mu, \sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}= \frac{5}{\sqrt{22}}=1.066)](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%5Csim%20N%28%5Cmu%2C%20%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B22%7D%7D%3D1.066%29)
And for this case we want this probability:
![P(64.3](https://tex.z-dn.net/?f=P%2864.3%3C%5Cbar%20X%3C66.4%29%3DP%28%5Cfrac%7B64.3-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3C%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3C%5Cfrac%7B64.3-%5Cmu%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%29%3DP%28%5Cfrac%7B64.3-65%7D%7B1.066%7D%3CZ%3C%5Cfrac%7B66.4-65%7D%7B1.066%7D%29%3DP%28-0.657%3Cz%3C1.313%29)
And we can find this probability on this way:
![P(-0.657](https://tex.z-dn.net/?f=P%28-0.657%3Cz%3C1.313%29%3DP%28z%3C1.313%29-P%28z%3C-0.657%29)
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
![P(-0.657](https://tex.z-dn.net/?f=P%28-0.657%3Cz%3C1.313%29%3DP%28z%3C1.313%29-P%28z%3C-0.657%29%3D0.905-0.256%3D0.650)
Part c
For this case since the sample mean is defined as:
![\bar x = \frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%20%5Cbar%20x%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
If we find the expected value for this variable we got:
![E(\bar X)= \frac{1}{n} \sum_{i=1}^n X_i =\frac{n\mu}{n}=\mu](https://tex.z-dn.net/?f=%20E%28%5Cbar%20X%29%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5En%20X_i%20%3D%5Cfrac%7Bn%5Cmu%7D%7Bn%7D%3D%5Cmu)
And for the variance we have:
![Var(\bar X) =\frac{1}{n^2} \sum_{i=1}^n Var(Xi) =\frac{n \sigma^2}{n^2}= \frac{\sigma^2}{n}](https://tex.z-dn.net/?f=%20Var%28%5Cbar%20X%29%20%3D%5Cfrac%7B1%7D%7Bn%5E2%7D%20%5Csum_%7Bi%3D1%7D%5En%20Var%28Xi%29%20%3D%5Cfrac%7Bn%20%5Csigma%5E2%7D%7Bn%5E2%7D%3D%20%5Cfrac%7B%5Csigma%5E2%7D%7Bn%7D)
And for this reason the deviation is ![\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
And since the distribution of X is normal we can conclude that the distribution for the sample mean is also normal and for this reason we can use the z table no matter if the sample size is small.