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:

$X \sim N(65.5,2.6)$

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(64.2$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(64.2$

And we can find this probability on this way:

$P(-0.50$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.50$

3) Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (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:

$P(X$