8.1: Normal Random Variables

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Section 1: Introduction

Definition: Normal probability density curve

A normal probability density curve with parameters $\mu$ and $\sigma$ is a bell-shaped curve that satisfies the following properties:

Has the peak at $\mu$ and symmetric about $\mu$.
Extends indefinitely in both directions, approaching but never touching the horizontal axis.
The Empirical rule holds that is:
1. ~68% of the area under the curve is between $\mu-1\sigma$ and $\mu+1\sigma$;
2. ~95% of the area under the curve is between $\mu-2\sigma$ and $\mu+2\sigma$;
3. ~99.7% of the area under the curve is between $\mu-3\sigma$ and $\mu+3\sigma$.

The description of the curve should sound familiar as we have seen it before!

$clipboard_ed38fa684f484ae0871f15fa0a2a5ac2a.png$

Definition: Normal random variable with parameters $\mu$ and $\sigma$

We define a normal random variable with parameters $\mu$ and $\sigma$ as a continuous random variable with a normal probability density curve with parameters $\mu$ and $\sigma$. We denote such a random variable in the following way:

$X \sim N(\mu, \sigma)$

Definition: Parameters of a normal distribution

The values, $\mu$ and $\sigma$, are called the parameters of a normal distribution and each pair of such values determines a unique shape of the probability density curve.

Example $\PageIndex{1.1}$

For example, when $\mu=50$ and $\sigma=10$ we get the following curve:

$clipboard_e6b766b9a2d428cb1db5ec588a43e1589.png$

Example $\PageIndex{1.2}$

For example, when $\mu=8$ and $\sigma=2$ we get the following curve:

$clipboard_ec7e2e4a5a0b6a30662b1387467bc76d1.png$

Let’s make sure we understand the properties of normal probability density curves and use the empirical rules to find some probabilities for any $X \sim N(\mu, \sigma)$.

To find the probability that $X$ is less than $\mu$ we will look for the area under the curve to the left of $\mu$. This region’s area can be found by adding the areas of included regions together to get $P(X<\mu)=0.50$ or $50\%$.
To find the probability that $X$ is between $\mu-\sigma$ and $\mu+\sigma$ we will look for the area under the curve between $\mu-\sigma$ and $\mu+\sigma$. This region’s area can be found by adding the areas of included regions together to get $P(\mu-\sigma<X<\mu+\sigma)=0.68$ or $68\%$.
To find the probability that $X$ is between $\mu$ and $\mu+2\sigma$ we will look for the area under the curve between $\mu$ and $\mu+2\sigma$. This region’s area can be found by adding the areas of included regions together to get $P(\mu<X<\mu+2\sigma)=0.475$ or $47.5\%$.
To find the probability that $X$ is between $\mu-\sigma$ and $\mu+2\sigma$ we will look for the area under the curve between $\mu-\sigma$ and $\mu+2\sigma$. This region’s area can be found by adding the areas of included regions together to get $P(\mu-\sigma<X<\mu+2\sigma)=0.815$ or $81.5\%$.
To find the probability that $X$ is greater than $\mu+\sigma$ we will look for the area under the curve to the right of . This region’s area can be found by adding the areas of included regions together to get $P(X>\mu+\sigma)=0.16$ or $16\%$.
To find the probability that $X$ is less than $\mu-2\sigma$ we will look for the area under the curve to the left of $\mu-2\sigma$. This region’s area can be found by adding the areas of included regions together to get $P(X<\mu-2\sigma)=0.025$ or $2.5\%$.
To find the probability that $X$ is greater than $\mu+4\sigma$ we will look for the area under the curve to the right of $\mu+4\sigma$. This region’s area is essentially equal to $0$ so the probability is nearly $0\%$, that is $P(X>\mu+4\sigma)=0$.

Example $\PageIndex{1.3}$

Example: For $X \sim N(\mu=50, \sigma=10)$, let’s find the following probabilities:

To find the probability that $X$ is less than $50$ we will look for the area under the curve to the left of $50$. This region’s area can be found by adding the areas of included regions together to get $P(X<50)=0.50$ or $50\%$.
To find the probability that $X$ is between $40$ and $70$ we will look for the area under the curve between $40$ and $70$. This region’s area can be found by adding the areas of included regions together to get $P(40<X<70)=0.815$ or $81.5\%$.

Example $\PageIndex{1.4}$

Example: For $X \sim N(\mu=8, \sigma=2)$, let’s find the following probabilities:

To find the probability that $X$ is between $8$ and $12$ we will look for the area under the curve between $8$ and $12$. This region’s area can be found by adding the areas of included regions together to get $P(8<X<12)=0.475$ or $47.5\%$.
To find the probability that $X$ is between $6$ and $12$ we will look for the area under the curve between $6$ and $12$. This region’s area can be found by adding the areas of included regions together to get $P(6<X<12)=0.815$ or $81.5\%$.

Section 2: Working with Normal Random Variables

For a normal random variable with parameters $\mu$ and $\sigma$, we want to be able to find the probabilities that involve the values $a$, $b$, and $c$ that are not covered by the Empirical rule:

$P(X<c)$
$P(X>c)$
$P(a<X<b)$

But first let’s make sure we understand the questions!

To find the probability that $X$ is less than $c$, we will look for the area under the curve to the left of $c$.
To find the probability that $X$ is greater than $c$, we will look for the area under the curve to the right of $c$.
To find the probability that $X$ is between $a$ and $b$, we will look for the area under the curve to the right of $a$ and to the left of $b$.

While it is clear what the regions’ areas look like they cannot be found by using the empirical rule. So how do we find the probabilities that involve values that are not covered by the Empirical rule such as $P(X<c)$, $P(X>c)$, $P(a<X<b)$.

For $X \sim N(\mu=8,\sigma=2)$, let's find the following probabilities:

$P(X<10.5)$
$P(X>9.3)$
$P(7.1<X<8.7)$

Luckily, there are ways to use technology to answer these questions:

$clipboard_ef4daafe32fb2d298dea0e4ea2c97e279.png$

$clipboard_e038ac03335d6a9049db0579e87f2b7ed.png$

$clipboard_e8543f848809d2f3bea30c79cc8cb27f3.png$

However, it is worth learning how to perform this task by hand. To be able to answer the same questions manually we have to review the concept of a z-score.

Definition: $z$-score

The $z$-score of an observation $x_i$ is computed by using the formula:

$z_{\text{score}}=\frac{x_i-\mu}{\sigma}$

Example $\PageIndex{2.1}$

Let’s find some $z$-scores for a normal random variable $X$ with $\mu=8$ and $\sigma=2$:

the z-score of $x_1=7.1$ is $\frac{7.1-8}{2}=-0.45$
the z-score of $x_2=8.7$ is ${8.7-8}{2}=0.35$
the z-score of $x_3=9.3$ is $\frac{9.3-8}{2}=0.65$
the z-score of $x_4=10.5$ is $\frac{10.5-8}{2}=1.25$

If instead of observing a normal variable with parameters $\mu$ and $\sigma$ we consider the $z$-scores of the observed values. Then it is not too hard to observe that the $z$-scores also have a normal distribution with parameters $0$ and $1$ which we call the standard normal distribution. In other words, the new random variable $\frac{X-\mu}{\sigma}$ has the standard normal distribution, therefore:

$\frac{X-\mu}{\sigma}=Z \sim N(0,1)$

By the way, the fact that the standard normal distribution is called $Z$-distribution explains the origin of the name for a term “a $z$-score”.

Definition: Standardizing a random variable

The process of converting the original normal random variable to $Z$ is called standardizing the random variable.
Most importantly, the area between any two values $a$ and $b$ under the normal curve is the same as the area between the $z$-scores of the $a$ and $b$ under the $z$-curve:

$P(a<X<b)=P(z_a<Z<z_b )$

Example $\PageIndex{2.2}$

For $X \sim N(\nu=8,\sigma=2)$, find the following probabilities:

$P(X<10.5)=P(Z<\frac{10.5-8}{2})=P(Z<1.25)=0.8944$
$P(X>9.3)=P(Z>\frac{9.3-8}{2})=P(Z>0.65)=0.2578$
$P(7.1<X<8.7)=P(\frac{7.1-8}{2}<Z<\frac{8.7-8}{2})=P(-0.45<Z<0.35)=P(Z<0.35)-P(Z<-0.45)=0.6368-0.3264=0.3104$

Similarly, we can use the standardization process for any other normal random variable.

Example $\PageIndex{2.3}$

For $X\ \sim N(\mu=50, \sigma=10)$, find the following probabilities:

$P(X<58)=P(Z<\frac{58-50}{10})=P(Z<0.8)=0.7881$
$P(X>32)=P(Z>\frac{32-50}{10})=P(Z>-1.8)=0.9641$
$P(47<X<69)=P(\frac{47-50}{10}<Z<\frac{69-50}{10})=P(-0.3<Z<1.9)=0.589$

Of course, every result obtained manually can be confirmed using technology:

$clipboard_e0e6c79ebfebfc4246c927d668d5991e3.png$

$clipboard_e2e130f953cc256e545562c2415ce4704.png$

$clipboard_e8868487fa775fe7430b1033bc4e93b5f.png$

Next, we will discuss how to find $x_\alpha$. To find $x_\alpha$ means to find the value the area to the right of which is equal to $\alpha$. In other words, we are looking for $x_\alpha$ such that

$P(X>x_\alpha)=\alpha$

Standardizing the left-hand side of the equation gives us

$P(X>x_\alpha )=P(Z>\frac{x_\alpha-\mu}{\sigma})=\alpha$

which implies that

$\frac{x_\alpha-\mu}{\sigma}=z_\alpha$

therefore

\(x_\alpha=\mu+z_\alpha\cdot\sigma

Example $\PageIndex{2.4}$

For a normal random variable with parameters $\mu=8$ and $\sigma=2$, let’s find $x_0.38$ by using the above formula.

$x_{0.38}=\mu+z_{0.38}\cdot\sigma$

Now that we related x_{0.38} to z_{0.38} we use the $z$-table as we have done it before to identify $\alpha=0.38$ and $1-\alpha=0.62$ to obtain the $z_{0.38}=0.31$. Now we can compute the desired $x_{0.38}$ by finishing the calculations:

$x_{0.38}=\mu+z_{0.38}\cdot\sigma=8+0.31⋅2=8.62$

Note that the following two statements are true by the definition of the alpha notation:

$P(X<8.62)=0.62$ and $P(X>8.62)=0.38$

Example $\PageIndex{2.5}$

For a normal random variable with parameters $\mu=50$ and $\sigma=10$, let’s find $x_{0.10}$ by using the above formula.

$x_{0.10}=\mu+z_{0.10}\cdot\sigma$

Now that we related $x_{0.10}$ to $z_{0.10}$ we use the $z$-table as we have done it before to identify $\alpha=0.10$ and $1-\alpha=0.90$ to obtain the $z_{0.10}=1.28$. Now we can compute the desired $x_{0.10}$ by finishing the calculations:

$x_{0.10}=\mu+z_{0.10}\cdot\sigma=50+1.28⋅10=62.8$

Now that we know how to find the probabilities and $x_\alpha$s let’s discuss the parameters of a normal distribution. It is not a coincidence that the two parameters, $\mu$ and $\sigma$, for a normal random variable are labeled similar to the mean and the standard deviation of the data set. The meaning and the interpretation of the mean and standard deviation for a continuous random variable is the same as for discrete random variables. That is, the expectation (or the mean) of a random variable is a measure of the center of the distribution and the variance (and standard deviation) is a measure of the spread of the random variable.

For $X \sim N(\mu, \sigma)$:

$E[N(\mu,\sigma)]=\mu$

$VAR[N(\mu,\sigma)]=\sigma^2$

$SD[N(\mu,\sigma)]=\sigma$

Example $\PageIndex{2.6}$

For example, for $X \sim N(10,2)$ by the three standard deviations rule, we may conclude that:

Range	Numerical Range	Interpretation
within $2\sigma_X$ from $\mu_X$	between 6 and 14	“expected”
more than $2\sigma_X$ from $\mu_X$	less than 6 or more than 14	“unusual”
more than $3\sigma_X$ from $\mu_X$	less than 4 or more than 16	“abnormal”

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Definition: Normal probability density curve

Definition: Normal random variable with parameters \(\mu\) and \(\sigma\)

Definition: Parameters of a normal distribution

Example \(\PageIndex{1.1}\)

Example \(\PageIndex{1.2}\)

Example \(\PageIndex{1.3}\)

Example \(\PageIndex{1.4}\)

Definition: \(z\)-score

Example \(\PageIndex{2.1}\)

Definition: Standardizing a random variable

Example \(\PageIndex{2.2}\)

Example \(\PageIndex{2.3}\)

Example \(\PageIndex{2.4}\)

Example \(\PageIndex{2.5}\)

Example \(\PageIndex{2.6}\)

Range	Numerical Range	Interpretation
within \(2\sigma_X\) from \(\mu_X\)	between 6 and 14	“expected”
more than \(2\sigma_X\) from \(\mu_X\)	less than 6 or more than 14	“unusual”
more than \(3\sigma_X\) from \(\mu_X\)	less than 4 or more than 16	“abnormal”