**Percentage of population greater than 80.****Percentage of population less than 90.****Percentage of population between 112 – 132.**

The question aims to find the **percentage** of the **people’s IQ** with the **mean** of the **population** to be 100 and a **standard deviation** of 16.

The question is based on the concepts of **probability** from a **normal distribution** using a z-table or z-score. It also depends on the **population’s mean** and the **population’s standard deviation.** The z-score is the **deviation** of a data point from the **population’s mean.** The formula for z-score is given as:

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

## Expert Answer

This question is based on the **normal model** which is given as:

\[ N(\mu, \sigma) = N(100, 16) \]

We can find the **percentage** of **population** for a given **limit** by using the $z-score$ which is given as follows:

**a)** The **percentage** of **population greater than** $X \gt 80$ can be calculated as:

\[ p = P(X \gt 80) \]

Converting the **limit** into $z-score$ as:

\[ p = P \big(Z \gt \dfrac{ 80\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \gt -1.25) \]

\[ p = 1\ -\ P(Z \lt -1.25) \]

Using the $z-$table, we get the $z-score$ of the above **probability** value to be:

\[ p = 1\ -\ 0.1056 \]

\[ p = 0.8944 \]

The **percentage** of **population** with **IQ** above $80$ is $89.44\%$.

**b)** The **percentage** of **population greater than** $X \lt 90$ can be calculated as:

\[ p = P(X \lt 90) \]

Converting the **limit** into $z-score$ as:

\[ p = P \big(Z \lt \dfrac{ 90\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \lt -0.625) \]

Using the $z-$table, we get the $z-score$ of the above **probability** value to be:

\[ p = 0.2660 \]

The **percentage** of **population** with **IQ** below $90$ is $26.60\%$.

**c)** The **percentage** of **population between** $X \gt 112$ and $X \lt 132$ can be calculated as:

\[ p = P(112 \lt X \lt 132 \]

Converting the **limit** into $z-score$ as:

\[ p = P \big(\dfrac{ 112\ -\ 100 }{ 16 } \lt Z \lt \dfrac{ 132\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \lt -2)\ -\ P(Z \lt 0.75) \]

Using the $z-$table, we get the $z-scores$ of the above **probability** values to be:

\[ p = 0.9772\ -\ 0.7734 \]

\[ p = 0.2038 \]

The **percentage** of **population** with **IQ** between $112$ and $132$ is $20.38\%$.

## Numerical Result

**a)** The **percentage** of **population** with **IQ** above $80$ is $89.44\%$.

**b) **The **percentage** of **population** with **IQ** below $90$ is $26.60\%$.

**c)** The **percentage** of **population** with **IQ** between $112$ and $132$ is $20.38\%$.

## Example

The **normal model** $N(55, 10)$ is given of people describing their **age.** Find the **percentage** of **people** with **age** below $60$.

\[ x = 60 \]

\[ p = P(X \lt 60) \]

\[ p = P \Big(Z \lt \dfrac{ 60\ -\ 55 }{ 10 } \Big) \]

\[ p = P(Z \lt 0.5) \]

\[ p = 0.6915 \]

The **percentage** of **people** with **age** below $60$ is $69.15\%$.