Changing the mean and standard deviation of a Normal distribution by a moderate amount can greatly change the percent of observations in the tails. Suppose that a college is looking for applicants with SAT math scores 750 and above.

a. In 2015, the scores of men on the math SAT followed the N ( 527 , 124 ) distribution. What percent of men scored 750 or better?

a. 96.41 %
b. 1.36 %
c. 0.96 %
d. 1.80 %
e. 3.51 %
f. 3.59 %

b. Women's SAT math scores that year had the N ( 496 , 115 ) distribution. What percent of women scored 750 or better?

a. 0.99 %
b. 1.50 %
c. 2.21 %
d. 1.36 %
e. 1.09 %
f. 1.45 %

a)

f. 3.59 %

b)

d. 1.36 %

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".  

Part a

Let X the random variable that represent the scores of men 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 complement rule and with the normal standard distribution table or excel:

So then the correct answer for this case would be:

f. 3.59 %

Part b

Let X the random variable that represent the scores of women's 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 complement rule and with the normal standard distribution table or excel:

So then the correct answer for this case would be:

d. 1.36 %

She does not have enough because if you add 188 and 258 for the total yards of fabric needed it would be 376. Mia has less than that, she would need 64 more yards of fabric and to find this you subtract the 312 from 376=64