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## Here's the Solution to this Question

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

$a)$

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.
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