Answered

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 :

Answer:

[tex]P(X<64)=P(\frac{X-\mu}{\sigma}<\frac{64-\mu}{\sigma})=P(Z<\frac{64-50}{8})=P(z<1.75)[/tex]

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

[tex]P(z<1.75)=0.9599[/tex]

Step-by-step 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:

[tex]X \sim N(50,8)[/tex]  

Where [tex]\mu=50[/tex] and [tex]\sigma=8[/tex]

We are interested on this probability

[tex]P(X<64)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X<64)=P(\frac{X-\mu}{\sigma}<\frac{64-\mu}{\sigma})=P(Z<\frac{64-50}{8})=P(z<1.75)[/tex]

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

[tex]P(z<1.75)=0.9599[/tex]

Other Questions