**-a) What percentage of 25-year-old men are over $6$ feet, $2$ inches tall?**

**-b) What percentage of men in the $6$-footer club are over $6$ feet, $5$ inches?**

This question aims to explain the **mean, variance, standard deviation,** and **z-score.**

The **mean** is the **central** or the most common **value** in a group of **numbers.** In statistics, it is a **measure** of the central trend of a **probability** distribution along **mode** and **median.** It is also **directed** as an expected **value.**

The term **variance** directs to a **statistical** stature of the **distribution** between **numerals** in a data set. More **precisely,** variance **estimates** how far each **numeral** in the set is from the **mean average,** and thus from every other **numeral** in the set. This **symbol:** $\sigma^2$ often expresses **variance.**

**Standard deviation** is a statistic that **estimates** the distribution of a **dataset** relative to its **mean** and is **calculated** as the square root of the **variance.** The standard deviation is **computed** as the square root of **variance** by defining each data point’s **deviation** as compared to the **mean.**

A **Z-score** is a numerical measure that defines a value’s connection to the mean of a **cluster** of values. Z-score is **calculated** in terms of standard **deviations** from the mean. If a **Z-score** is $0$, it indicates that the data point’s score is **similar** to the mean **score.**

## Expert Answer

Given the **mean** $\mu$ and the **variance,** $\sigma^2$ of a $25$-year **man** is $71$ and $6.25$,Â **respectively.**

**Part a **

To find the **percentage** of $25$-year-old men that are over $6$ feet and $2$ inches we first **calculate** the **probability** of $P[X> 6 feet \space 2 \space inches]$.

$6$ feet and $2$ inches can be **written** as $74 \space in$.

We have to find the $P[X>74 \space in]$ and it is **given** as:

\[P[X>74]=P\left[\dfrac{X-\mu}{\sigma}>\dfrac{74-71}{2.5}\right]\]

That is:

\[=P[Z\leq 1.2] \]

\[1-\phi(1.2) \]

\[1-0.8849\]

\[0.1151\]

**Part b**

In this **part,** we have to find the **height** of a $25$-year-old man **above** $6$ feet $5$ inches **given** that he is $6$ feet.

$6$ feet and $5$ inches can be **written** as $77 \space in$.

We have to **find** the $P[X>77 \space in | 72 \space in]$ and it is **given** as:

\[ P[X>77 \space in | 72 \space in] = \dfrac{X>77 | X>72}{P[X>72]} \]

\[= \dfrac{P[X>77]}{P[X>2]} \]

\[= \dfrac{ P \left[ \dfrac{X-\mu}{\sigma} > \dfrac{77-71}{2.5} \right]} {P \left[ \dfrac{X-\mu}{\sigma} > \dfrac{72-71}{2.5} \right] } \]

\[ \dfrac{P[Z >2.4]}{P[Z >0.4]} \]

\[ \dfrac{1- P[Z >2.4]}{P[Z >0.4]} \]

\[ \dfrac{1- 0.9918}{1- 0.6554} \]

\[ \dfrac{0.0082}{0.3446} \]

\[ 0.0024\]

## Numerical Results

**Part a: **The **percentage** of **men** above $6$ feet and $2$ inches is $11.5 \%$.

**Part b: **The **percentage** of 25-year-old men in the $6$-footer **club** that are **above** $6$ feet and $5$ inches is $2.4 \%$.

## Example

The **grades** on a math **final** at school have a **mean** $\mu = 85$ and a **standard** deviation of $\sigma = 2$. **John** scored $86$ on the exam. Find the **z-score** for John’s exam grade.

\[z=\dfrac{X-\mu}{\sigma}\]

\[z=\dfrac{86-85}{2}\]

\[z=\dfrac{1}{2}\]

\[z=0.5\]

John’s **z-score** is $0.5$.