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__ 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.

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.

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

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

**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 \(x=35.4\).

- Last Updated: Mar 29, 2023 11:00 AM
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