Question

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 %

Answer 1

Answer:

a) $P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z1.798)$

$P(Z>1.798)=1-P(z$

f. 3.59 %

b) $P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z2.21)$

$P(Z>2.21)=1-P(Z$

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:

$X \sim N(527,124)$  

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(X>750)$

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

$z=\frac{x-\mu}{\sigma}$

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

$P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z1.798)$

And we can find this probability using the complement rule and with the normal standard distribution table or excel:

$P(Z>1.798)=1-P(Z$

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:

$X \sim N(496,115)$  

Where $\mu=496$ and $\sigma=115$

We are interested on this probability

$P(X>750)$

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

$z=\frac{x-\mu}{\sigma}$

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

$P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z2.21)$

And we can find this probability using the complement rule and with the normal standard distribution table or excel:

$P(Z>2.21)=1-P(Z$

So then the correct answer for this case would be:

d. 1.36 %