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: \[f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{x-\mu}{\sigma} \right]^2}\] 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.
Example: Find the probability that a standard normal random variable lies
See the video below for the solution
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 \[z=\frac{x-\mu}{\sigma}\] 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 \(x_0\) 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 \(x=35.4\).