Answer:
a) ![P(64.2](https://tex.z-dn.net/?f=P%2864.2%3CX%3C67.2%29%3DP%28-0.50%3Cz%3C0.65%29%3DP%28z%3C0.65%29-P%28-0.5%29%3D0.742-0.309%3D0.434)
b) 69.764
Step-by-step explanation:
1) 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".
2) Part a
Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:
Where
and ![\sigma=2.6](https://tex.z-dn.net/?f=%5Csigma%3D2.6)
We are interested on this probability
![P(64.2](https://tex.z-dn.net/?f=P%2864.2%3CX%3C67.2%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(64.2](https://tex.z-dn.net/?f=P%2864.2%3CX%3C67.2%29%3DP%28%5Cfrac%7B64.2-%5Cmu%7D%7B%5Csigma%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3C%5Cfrac%7B67.2-%5Cmu%7D%7B%5Csigma%7D%29%3DP%28%5Cfrac%7B64.2-65.5%7D%7B2.6%7D%3CZ%3C%5Cfrac%7B67.2-65.5%7D%7B2.6%7D%29%3DP%28-0.50%3Cz%3C0.65%29)
And we can find this probability on this way:
![P(-0.50](https://tex.z-dn.net/?f=P%28-0.50%3Cz%3C0.65%29%3DP%28z%3C0.65%29-P%28-0.5%29)
And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.
![P(-0.50](https://tex.z-dn.net/?f=P%28-0.50%3Cz%3C0.65%29%3DP%28z%3C0.65%29-P%28-0.5%29%3D0.742-0.309%3D0.434)
3) Part b
For this part we want to find a value a, such that we satisfy this condition:
(a)
(b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05
If we use condition (b) from previous we have this:
![P(z](https://tex.z-dn.net/?f=P%28z%3C%5Cfrac%7Ba-%5Cmu%7D%7B%5Csigma%7D%29%3D0.95)
But we know which value of z satisfy the previous equation so then we can do this:
![z=1.64](https://tex.z-dn.net/?f=z%3D1.64%3C%5Cfrac%7Ba-65.5%7D%7B2.6%7D)
And if we solve for a we got
![a=65.5 +1.64*2.6=69.764](https://tex.z-dn.net/?f=a%3D65.5%20%2B1.64%2A2.6%3D69.764)
So the value of height that separates the bottom 95% of data from the top 5% is 69.764.