We are given a population mean `mu=.260` and a standard deviation `sigma=.07` :

(1) What is the probability of a random individual hitting would hit better than .314?

Assuming that batting averages are approximately normally distributed:

First compute the corresponding `z` value using `z=(x-mu)/sigma`

`z=(.314-.260)/.07~~.7714`

Now `P(x>= .314)=P(z>=.7714)`

Consulting a standard normal table we find that the area to the right of z=.7714 (and therefore the probability that a random z score would be in this area) is approximately 0.22.

**Thus the probability of a random player hitting greater than .314 is approximately 22%.**

(2) To be in the 99th percentile: We consult the standard normal table to find the z-score associated with an area to the left of .9900

The z-score is 2.33

The associated batting avereage can be found using the formula for the z-score:

`2.33=(x-.26)/.07`

`x-.26=.1631`

`x=.4231`

**So a batter would have to be hitting .423 or better to be in the 99th percentile.**