Question

The heights of 100 students are normally distributed with a mean height of 16 centimeters and

standard deviation of 5.9 centimeters. Answer the following in percent form.

a. How many of these students would you expect to have heights less than 167 centimeters?

b. How many of these students would you expect to have heights greater than 153 centimeters?

c. How many of these students would you expect to have heights between 150 and 165

centimeters?

d. Find the area of items a, b and c.

e. Compute for the 75th Percentile.

Answer 1

$n=100\\\mu=160\\\sigma=5.9$

$a)$

To find the expected number of students who have heights less than 167 centimeters, we first determine the probability,

$p(x\lt 167)=p(Z\lt{167-160\over5.9})=p(Z\lt1.19)=0.8830$

To convert this probability into percent form, we multiply by 100%. So we have, $0.8830\times100\%=88.30\%$

Therefore, we would expect 88.30% of the student's height to be less than 167 centimeters.

$b)$

To determine the expected number of students with heights greater than 153 centimeters, we first the probability,

$p(x\gt153)=p(Z\gt {153-160\over 5.9})=p(Z\gt -1.19)=1-p(Z\lt-1.19)=1-0.1170=0.8830$In percent form, $0.8830\times100\%=88.30\%$ .Therefore, we would expect that the heights of 88.30% of the students would be greater than 153 centimeters.

$c)$  

We find the probability,

$p(150\lt x\lt165)=p({150-160\over5.9}\lt Z\lt {165-160\over 5.9})=p(-1.69\lt Z\lt 0.85)$

We can write this as,

$p(-1.69\lt Z\lt 0.85)=\phi(0.85)-\phi(-1.69)=0.8023-0.0455=0.7568$

In percentage form, .

$0.7568\times 100\%=75.68\%$

Therefore, we would expect that 75.68% of the students have heights between 150 and 165 centimeters.

$d)$

The area for the items a, b and c above are just their respective probability.

$i)$

For part a, we determine,

$p(x\lt 167)=p(Z\lt{167-160\over5.9})=p(Z\lt1.19)=0.8830$

$ii)$

For part b, we find,

$p(x\gt153)=p(Z\gt {153-160\over 5.9})=p(Z\gt -1.19)=1-p(Z\lt-1.19)=1-0.1170=0.8830$

$iii)$

For part c, we find,

$p(150\lt x\lt165)=p({150-160\over5.9}\lt Z\lt {165-160\over 5.9})=p(-1.69\lt Z\lt 0.85)=p(Z\lt 0.85)-p(Z\lt -1.69)=0.8023-0.0455=0.7568$

$e)$

We can find the $75^{th}$ percentile by determining a value $y$ such that,

$p(x\lt y)=0.75$  

So,

$p(x\lt y)=p(Z\lt{y-160\over5.9})=0.75$

This implies that,

${y-160\over5.9}=Z_{0.75}$ where $Z_{0.75}$ is the standard normal table value associated with 0.75. The value of $Z_{0.75}=0.674$

This implies that,

${y-160\over5.9}=0.674\implies y=160+3.9766=163.9766\approx. 164$

Therefore, the $75^{th}$ percentile is 164 centimeters.