Question

Standardization of a Normal Distribution: Bryce reads in the latest issue of Pigskin Roundup that the average number of rushing yards per game by NCAA Division II starting running backs is 50 with a standard deviation of 8 yards. If the number of yards per game (X) is normally distributed, what is the probability that a randomly selected running back has 64 or fewer rushing yards

Answer 1

`P(X<64)=P((X-\mu)/(\sigma)<(64-\mu)/(\sigma))=P(Z<(64-50)/(8))=P(z<1.75)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<1.75)=0.9599`

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".

Solution to the problem

Let X the random variable that represent the number of rushing yards of a population, and for this case we know the distribution for X is given by:

`X \sim N(50,8)`

Where
`\mu=50` and
`\sigma=8`

We are interested on this probability

`P(X<64)`

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

`z=(x-\mu)/(\sigma)`

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

`P(X<64)=P((X-\mu)/(\sigma)<(64-\mu)/(\sigma))=P(Z<(64-50)/(8))=P(z<1.75)`

And we can find this probability using the normal standard table or excel and we got:

`P(z<1.75)=0.9599`