Skip to Main Content


The Normal Distribution

A common random continuous distribution you will encounter in statistics is the normal distribution. It is bell-shaped and it is sometimes called the bell-curve. It is a continuous probability distribution relating to the mean and standard deviation. 


The normal distribution plays an important role in inferential statistics. 

Probability Distribution for a Normal Random Variable x

Probability density function: 

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


\(\mu =\) Population mean of the normal random variable \(x\)

\(\sigma =\) Population standard deviation

\(P(x<a)\) is obtained from a table of normal probabilities.

The standard normal distribution is a normal distribution with \(\mu =0\) and \(\sigma =1\). A random variable with a standard normal distribution, denoted by the symbol \(z\), is called a standard normal variable.

Instead of plugging into the function above to find each probability, we can use a table of z variables to find the probability.

Using the Z-Table

Example: Find the probability that a standard normal random variable lies

  1. to the left of 1.23.
  2. to the right of 1.23.
  3. in between -0.26 and 1.23.

See the video below for the solution

Converting a Normal Distribution to a Standard Normal Distribution

The z-table refers to the standard normal distribution with mean \(\mu=0\) and standard deviation \(\sigma=1\). To apply normal distributions with different means or standard deviations we have to convert the value of \(x\) to a z-score before looking up the table.

Converting a Normal Distribution to a Standard Normal Distribution

If \(x\) is a normal random variable with mean \(\mu\) and standard deviation \(\sigma\), then the random variable \(z\) defined by the formula 


has a standard normal distribution. The value \(z\) describes the number of standard deviations between \(x\) and \(\mu\).

Example 1: Suppose \(x\) is a normally distributed random variable with \(\mu=11\) and \(\sigma=2\). Find the probability that \(x\) is between 7.8 and 12.6.


First, we have to convert the \(x\) values 7.8 and 12.6 into z-scores.

\[z_1=\frac{7.8-11}{2} =-1.6\]

\[z_2=\frac{12.6-11}{2} =0.8\]

So, \(P(7.8<x<12.6)=P(-1.6<z<0.8)\)

Next, we look up the probabilities to the left of \(z_1=-1.6\) and \(z_2=0.8\) and find the difference.

From z-table, \(P(z<-1.6)= 0.0548\) and \(P(z<0.8)=0.7881\).

\begin{align} P(7.8<x<12.6)&=P(-1.6<z<0.8) \\  &= 0.7881 - 0.0548\\ &=0.7333\end{align}

Example 2: Suppose \(x\) is a normally distributed random variable with \(mu=30\) and \(\sigma=8\). Find a value \(x_0\) of the random variable \(x\) such that 

  1. \(P(x<x_0)=0.8\)
  2. 25% of the values are greater than \(x_0\)


1. To find 0.8 or 80% of the population, we have to look inside the z-table.


We can say 80% or 0.8000 is close to the value of 0.7995. 0.7995 represents the z-score 0.84.

\[P(z<0.84)=0.7995 \approx 0.8\]

We now need to find the \(x_0\) that corresponds to \(z=0.84\) using the formula,

\begin{align} z&=\frac{x-\mu}{\sigma}\\ x&=\mu+z\sigma\\ x&=30+0.84(8)=36.72\end{align}

Therefore, approximately 80% of the data is less than \(x=36.72\).

2. To find \(x_0\) value representing the largest 25% of the data, we once again use the z-table. But since the z-table represents values less than, we are looking up the the value that represents 75% or 0.7500 of the data. 


Since 75% or 0.7500 is almost exactly between the z-scores 0.67 and 0.68. We find the midpoint and say it is approximately equal to 0.675.

\[P(z<0.675) \approx 0.75\]

Since we are looking for the largest 25%, we change the relation around to \(P(z>0.675) \) and find the \(x\) value.

\begin{align} x&=\mu+z\sigma\\ x&=30+0.675(8)=35.4\end{align}

Therefore, approximately 25% of the data is greater than \(x=35.4\).

Statistics by Matthew Cheung. This work is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License.

chat loading...