# What Is a Normal Distribution in Statistics?

The normal distribution is the most important probability distribution.

Theory

### NormalDistribution

The normal distribution with expectation value $\mu$ and standard deviation $\sigma$ is described by the function

 $f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma }^{2}}}.$

Luckily, you will not be working with this function directly.

The probability of getting a result between $a$ and $b$ is given by the area bounded by the graph, the $x$-axis and vertical lines at $x=a$ and $x=b$. $P\phantom{\rule{-0.17em}{0ex}}\left(a\ge x\ge b\right)$ can be found in a probability table or through the use of digital tools.

When you’re working with the normal distribution by hand, you need to convert your normal distribution into the standard normal distribution. Then you can use the standardized table that contains all the solutions. A standardized normal distribution is a normal distribution with an expected value equal to 0 and a standard deviation equal to 1. When converting into the standard normal distribution, you use the following conversion formula:

Formula

### ConversiontoStandardNormalDistribution

 $Z=\frac{X-\mu }{\sigma },$

where the random variable $Z$ is distributed just like the standard normal distribution.

Example 1

The birth weight of a newborn girl is considered normally distributed with an expectation value of $\text{}3.50\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$ and a standard deviation of $\text{}0.48\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$.

1.
What is the probability that a random girl weighs less than $\text{}2.5\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$ when born?
2.
What is the probability that a random girl weighs more than $\text{}4.0\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$ when born?
3.
What is the probability that a random girl weighs between $\text{}2.5\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$ and $\text{}4.0\text{}\phantom{\rule{0.17em}{0ex}}\text{kg}$ when born?

1.
You want to find $P\phantom{\rule{-0.17em}{0ex}}\left(X\le 2.5\right)$. The calculation looks like this: $\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X\le 2.5\right)& =P\phantom{\rule{-0.17em}{0ex}}\left(\frac{X-\mu }{\sigma }\le \frac{2.5-3.5}{0.48}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\phantom{\rule{-0.17em}{0ex}}\left(Z\le -2.0833\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You look up the $Z$-value in the probability table for the normal distribution and find that

$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(Z\le -2.0833\right)⇔Z& =0.0188\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1.88\phantom{\rule{0.17em}{0ex}}\text{%}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{lll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(Z\le -2.0833\right)⇔Z=0.0188=1.88\phantom{\rule{0.17em}{0ex}}\text{%}.& \phantom{\rule{2em}{0ex}}& \hfill \end{array}$

The probability of a new born girl weighing less than $2.5$ kg is $1.88$ %.
2.
You want to find $P\phantom{\rule{-0.17em}{0ex}}\left(X\ge 4.0\right)$. The calculation looks like this: $\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(X\ge 4.0\right)& =1-P\phantom{\rule{-0.17em}{0ex}}\left(X\le 4.0\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1-P\phantom{\rule{-0.17em}{0ex}}\left(\frac{X-\mu }{\sigma }\le \frac{4.0-3.5}{0.48}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1-P\phantom{\rule{-0.17em}{0ex}}\left(Z\le 1.04\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You look up the $Z$-value in the table and find that

 $P\phantom{\rule{-0.17em}{0ex}}\left(Z\le 1.04\right)=0.8508$

which gives you that $\begin{array}{llll}\hfill 1-P\phantom{\rule{-0.17em}{0ex}}\left(Z\le 1.04\right)& =1-0.8508\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.1492\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =14.92\phantom{\rule{0.17em}{0ex}}\text{%}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The probability that a random newborn girl weighs more than $4.0$ kg is $14.92$ %.

3.
You want to find $P\phantom{\rule{-0.17em}{0ex}}\left(2.5\le X\le 4.0\right)$. In this case, you have to set up a double inequality. It looks like this:
$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(2.5\le X\le 4.0\right)& =P\left(\frac{2.5-3.5}{0.48}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{X-\mu }{\sigma }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\le \frac{4.0-3.5}{0.48}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\left(-2.08\le Z\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\le 1.04\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\phantom{\rule{-0.17em}{0ex}}\left(Z\le 1.04\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}-P\phantom{\rule{-0.17em}{0ex}}\left(Z\le -2.08\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(2.5\le X\le 4.0\right)& =P\phantom{\rule{-0.17em}{0ex}}\left(\frac{2.5-3.5}{0.48}\le \frac{X-\mu }{\sigma }\le \frac{4.0-3.5}{0.48}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\phantom{\rule{-0.17em}{0ex}}\left(-2.08\le Z\le 1.04\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =P\phantom{\rule{-0.17em}{0ex}}\left(Z\le 1.04\right)-P\phantom{\rule{-0.17em}{0ex}}\left(Z\le -2.08\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

From the table you find the values of the respective $Z$-values, and their difference is $\begin{array}{llll}\hfill P\phantom{\rule{-0.17em}{0ex}}\left(2.5\le X\le 4.0\right)& =0.8508-0.0188\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =0.8320\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =83.2\phantom{\rule{0.17em}{0ex}}\text{%}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The probability that a random newborn girl weighs between $2.5$ kg and $4.0$ kg is $83.2$ %.