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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 %

Respuesta :

Answer:

a) [tex]P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z<\frac{750-527}{124})=P(Z>1.798)[/tex]

[tex]P(Z>1.798)=1-P(z<1.798)= 1-0.9641 =0.0359 [/tex]

f. 3.59 %

b) [tex]P(X>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z<\frac{750-496}{115})=P(Z>2.21)[/tex]

[tex]P(Z>2.21)=1-P(Z<2.21)= 1-0.9864 =0.0136[/tex]

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:

[tex]X \sim N(527,124)[/tex]  

Where [tex]\mu=65.5[/tex] and [tex]\sigma=2.6[/tex]

We are interested on this probability

[tex]P(X>750)[/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>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z<\frac{750-527}{124})=P(Z>1.798)[/tex]

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

[tex]P(Z>1.798)=1-P(Z<1.798)= 1-0.9641 =0.0359[/tex]

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:

[tex]X \sim N(496,115)[/tex]  

Where [tex]\mu=496[/tex] and [tex]\sigma=115[/tex]

We are interested on this probability

[tex]P(X>750)[/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>750)=P(\frac{X-\mu}{\sigma}>\frac{750-\mu}{\sigma})=P(Z<\frac{750-496}{115})=P(Z>2.21)[/tex]

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

[tex]P(Z>2.21)=1-P(Z<2.21)= 1-0.9864 =0.0136[/tex]

So then the correct answer for this case would be:

d. 1.36 %

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