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 $.