It was estimated that batting averages in major league basketball in the 1940s were normally distributed with a mean of 0.260 and a standard......deviation of 0.07. a) What is the probability that...
It was estimated that batting averages in major league basketball in the 1940s were normally distributed with a mean of 0.260 and a standard......
deviation of 0.07.
a) What is the probability that a randomly selected batter would bat over 314, that is, their average would be greater than 0.314?
b) what batting average would a batter need to be in the 99th percentile?
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`
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:
So a batter would have to be hitting .423 or better to be in the 99th percentile.