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

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

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:

Where and

We are interested on this probability

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

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

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

answdf

step-by-step explanation: bcuz d is right

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2