Answer :
Answer:
(a) The height of 70.90 inches represents the 95th percentile.
(b) The height of 64.27 inches represents the first quartile.
Step-by-step explanation:
We are given that in a survey of women in a certain country (ages 20-29), the mean height was 66.2 inches with a standard deviation of 2.86 inches.
Let X = heights of women in a certain country.
So, X ~ Normal([tex]\mu=66.2,\sigma^{2} =2.86^{2}[/tex])
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean height = 66.2 inches
[tex]\sigma[/tex] = standard deviation = 2.86 inches
(a) We have to find the height that represents 95th percentile, that means;
P(X < x) = 0.95 {wherex is the required height}
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-66.2}{2.86}[/tex] ) = 0.95
P(Z < [tex]\frac{x-66.2}{2.86}[/tex] ) = 0.95
Now, in the z table the critical value of x that represents the top 5% area is given by 1.645, that is;
[tex]\frac{x-66.2}{2.86}=1.645[/tex]
[tex]{x-66.2=1.645\times 2.86[/tex]
x = 66.2 + 4.70 = 70.90 inches
So, the height of 70.90 inches represents the 95th percentile.
(b) We have to find the height that represents the first quartile, that means;
P(X < x) = 0.25 {wherex is the required height}
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-66.2}{2.86}[/tex] ) = 0.25
P(Z < [tex]\frac{x-66.2}{2.86}[/tex] ) = 0.25
Now, in the z table the critical value of x that represents the below 25% area is given by -0.6745, that is;
[tex]\frac{x-66.2}{2.86}=-0.6745[/tex]
[tex]{x-66.2=-0.6745\times 2.86[/tex]
x = 66.2 - 1.93 = 64.27 inches
So, the height of 64.27 inches represents the first quartile.