This **article aims to solve a normal random variable** **X** with $ \mu = 10$ and $ \sigma ^ {2} = 36$. This article uses the** normal random variable** concept. Like the **standard normal distribution**, all normal distributions are **unimodal** and **symmetrically distributed** with a **bell-shaped curve.** However, the **normal distribution** can take any value as its **mean** and **standard deviation**. **Mean** and **standard deviation** are always fixed in the standard normal distribution.

Each **normal distribution** is a version of the standard normal distribution that has been** stretched or squashed** and **shifted horizontally to the right or left. **The diameter determines where the **center of the curve** is. **Increasing** the diameter shifts the curve to the right, and **decreasing** it shifts the **curve to the left. **The **standard deviation** stretches or **compresses the curve**.

**Expert Answer**

Given $ X $ is the **normal random variable** with $ \mu = 10 $ and $ \sigma ^{2} = 36 $.

To **compute the following probabilities**, we will make use of the fact of $ X \sim N (\mu, \sigma ^{2} ) $, then $Z=\dfrac { X – \mu}{ \sigma } \sim N (0,1) $.

$ Z $ is the **standard normal variable** $ \Phi $ is its **CDF, whose probabilities** can be computed using the **standard normal table.**

\[ P [ X < 20 ] = P [ \dfrac { X- \mu }{ \sigma } < \dfrac { 20 – 10 }{ 6 }]\]

\[ = P [Z < \dfrac { 5 }{ 3 }] \]

\[ = \Phi (\dfrac { 5 } { 3 })\]

\[ = 0.9522 \]

**Numerical Result**

The** output of the expression** $ P [X < 20 ] $ with $ \mu = 10 $ and $ \sigma ^ {2} = 36 $ is $ 0.9522 $.

**Example**

**Given that $ X $ is a normal random variable with parameters $ \mu = 15 $ and $ \sigma ^ {2} = 64 $, compute $ P [X < 25] $.**

**Solution**

Given $ X $ is the **normal random variable** with $ \mu = 15 $ and $ \sigma ^{2} = 64 $.

To **compute the following probabilities**, we will make use of the fact of $ X \sim N (\mu, \sigma ^{ 2 } ) $, then $ Z = \dfrac { X – \mu }{ \sigma } \sim N (0,1) $.

$ Z $ is the **standard normal variable** $ \Phi $ is its **CDF, whose probabilities** can be computed using the **standard normal table.**

\[ P [ X < 25 ] = P [ \dfrac { X- \mu }{ \sigma } < \dfrac { 25 – 15 }{ 8 } ]\]

\[ =P [ Z < \dfrac {10 }{ 8 } ] \]

\[ = \Phi (\dfrac { 5 } { 4 })\]

\[ = 0.89435 \]

The** output of the expression** $ P [X < 25 ]$ with $ \mu = 15 $ and $ \sigma ^ { 2 } = 64 $ is $ 0.89435 $.