In a survey of women in a certain country (ages 20minus29), the mean height was 66.7 inches with a standard deviation of 2.77 inches. Answer the following questions about the specified normal distribution.a) What height represents the 99th percentile? (b) What height represtents the firsth quartile?

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 heights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(66.7,2.77)[/tex]

Where [tex]\mu=66.7[/tex] and [tex]\sigma=2.77[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.01[/tex] (a)

[tex]P(X<a)=0.99[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.99 of the area on the left and 0.01 of the area on the right it's z=2.33. On this case P(Z<2.33)=0.99 and P(z>2.33)=0.01

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.99[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.99[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=2.33<\frac{a-66.7}{2.77}[/tex]

And if we solve for a we got

[tex]a=66.7 +2.33*2.77=73.15[/tex]

So the value of height that separates the bottom 99% of data from the top 1% is 73.15.

Part b

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.75[/tex] (a)

[tex]P(X<a)=0.25[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-0.674<\frac{a-66.7}{2.77}[/tex]

And if we solve for a we got

[tex]a=66.7 -0.674*2.77=64.833[/tex]

So the value of height that separates the bottom 25% of data from the top 75% is 64.833.