Unless otherwise stated, the material in this guide is from The Learning Centre at Centennial College. Content has been adapted for the NWP Learning Commons in March 2022. This work is licensed under a Creative Commons BY 4.0 International License.
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.
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Probability Distribution for a Normal Random Variable x Probability density function: \[f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{x-\mu}{\sigma} \right]^2}\] where \(\mu =\) 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.
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.
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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 \[z=\frac{x-\mu}{\sigma}\] 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.
Solution
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
Solution
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 the \(x_0\) 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 \(x=35.4\).