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# Statistics

## 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:  $f(x)=\frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{1}{2}\left[\frac{x-\mu}{\sigma} \right]^2}$ where $$\mu =$$ Population mean of the normal random variable $$x$$ $$\sigma =$$ Population standard deviation $$P(x  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  $z=\frac{x-\mu}{\sigma}$ 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

1. $$P(x<x_0)=0.8$$
2. 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 $$x_0$$ 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 $$x=35.4$$.