Expected Value of Normal Distribution

The Normal Mean Lies at the Center of Symmetry

The expected value is the probability-weighted average of a random variable, or the long-run average approached across many independent observations. It is not generally the same as the most frequent value. For a normal distribution, however, the mean, median, mode, and expected value all meet at the same center.

This symmetry gives the normal distribution a simple expected-value rule.

If $X \sim N(\mu, \sigma^2)$ with $\sigma>0$ , its expected value equals the mean parameter $\mu$ .

Mathematically, we can write:

E(X) = \mu

In this notation, $X$ is a normally distributed random variable and $\mu$ is the distribution's mean parameter.

The following calculation shows why.

Deriving the Expected Value from the Integral

The expected value of a continuous random variable is defined as:

E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx

For a normal distribution with $\sigma>0$ , the probability density function is:

f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

Substitute this function into the expected value formula:

E(X) = \int_{-\infty}^{\infty} x \cdot \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \, dx

Use the substitution $t = \frac{x-\mu}{\sigma}$ . Then $x = \sigma t + \mu$ and $dx = \sigma \, dt$ .

Write the original density after substituting $x=\sigma t+\mu$ as $h(t)$ , and keep $dx=\sigma\,dt$ separate:

h(t)=\frac1{\sigma\sqrt{2\pi}}e^{-\frac12t^2}

Evaluate the two integrals separately:

First integral $\int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt = 0$ because $g(t)=t e^{-\frac{1}{2}t^2}$ is odd. Thus $g(-t) = -g(t)$ , and the signed areas over the symmetric interval $(-\infty, \infty)$ cancel.
Second integral $\int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt = \sqrt{2\pi}$ is the Gaussian integral. After multiplication by the normalizing factor $1/\sqrt{2\pi}$ , it gives total probability $1$ for the standard normal density.

The calculation gives:

E(X) = \frac{\sigma}{\sqrt{2\pi}} \cdot 0 + \frac{\mu}{\sqrt{2\pi}} \cdot \sqrt{2\pi}

The expected value of a normal random variable equals its mean parameter.

For every normal distribution $X \sim N(\mu, \sigma^2)$ , we have $E(X) = \mu$ . This means we don't need to perform integration every time we calculate the expected value of a normal distribution. We simply use the parameter value $\mu$ .

Practical Interpretation

For a normal distribution, the expected value and mean parameter are related as follows.

For a normal random variable with mean $\mu$ and standard deviation $\sigma$ , the expected value is exactly $\mu$ . Across many independent observations from this distribution, the sample average approaches $\mu$ under the law of large numbers.

Example:

If the heights of a group of students are normally distributed with mean $165 \text{ cm}$ , then the expected height of a randomly selected student is $165 \text{ cm}$ .

Expected Values for Exam Scores and Birth Weights

Example 1

Suppose the math exam scores in a class are normally distributed with mean $\mu = 75$ and standard deviation $\sigma = 10$ . What is the expected value of a student's exam score?

Solution:

Since normal distribution has the property $E(X) = \mu$ , the expected value of the exam score is:

E(X) = 75

The expected value of a student's exam score is $75$ .

Example 2

Suppose newborn weights at a hospital are modeled by a normal distribution with mean $3{,}200 \text{ grams}$ and standard deviation $400 \text{ grams}$ . Determine the expected weight of one randomly selected newborn.

Solution:

Given $\mu = 3{,}200 \text{ grams}$ and $\sigma = 400 \text{ grams}$ .

The expected weight is the mean parameter:

E(X) = \mu = 3{,}200 \text{ grams}

Exercises

Both problems give the mean and the standard deviation of a normal distribution and ask for the expected value. Read the mean directly from the description, because the expected value of a normal distribution equals its center.

Travel time from home to school is modeled by a normal distribution with mean $25 \text{ minutes}$ and standard deviation $5 \text{ minutes}$ . Determine the expected travel time.
Daily air temperature in Jakarta during June is modeled by a normal distribution with mean $28^\circ\text{C}$ and standard deviation $3^\circ\text{C}$ . What is the expected temperature on a randomly selected day?
Physics exam scores of grade $12$ students are modeled by a normal distribution with mean $78$ and standard deviation $12$ . What is the expected score of a randomly selected student?

Answer Key

Solution to Problem 1:

Given: $\mu = 25 \text{ minutes}$ , $\sigma = 5 \text{ minutes}$

Since normal distribution has the property $E(X) = \mu$ , then:

$E(X) = 25 \text{ minutes}$

Answer: The expected value of travel time is $25 \text{ minutes}$ .
Solution to Problem 2:

Given: $\mu = 28^\circ\text{C}$ , $\sigma = 3^\circ\text{C}$

Using the basic property of normal distribution:

$E(X) = \mu = 28^\circ\text{C}$

Answer: The expected value of air temperature is $28^\circ\text{C}$ .
Solution to Problem 3:

Given: $\mu = 78$ , $\sigma = 12$

A normally distributed variable satisfies $E(X)=\mu$ . Therefore:

$E(X) = \mu = 78$

Answer: The expected exam score is $78$ .

The Normal Mean Lies at the Center of Symmetry

This symmetry gives the normal distribution a simple expected-value rule.

If $X \sim N(\mu, \sigma^2)$ with $\sigma>0$ , its expected value equals the mean parameter $\mu$ .

Mathematically, we can write:

E(X) = \mu

In this notation, $X$ is a normally distributed random variable and $\mu$ is the distribution's mean parameter.

The following calculation shows why.

Deriving the Expected Value from the Integral

The expected value of a continuous random variable is defined as:

E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx

For a normal distribution with $\sigma>0$ , the probability density function is:

f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

Substitute this function into the expected value formula:

E(X) = \int_{-\infty}^{\infty} x \cdot \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \, dx

Use the substitution $t = \frac{x-\mu}{\sigma}$ . Then $x = \sigma t + \mu$ and $dx = \sigma \, dt$ .

Write the original density after substituting $x=\sigma t+\mu$ as $h(t)$ , and keep $dx=\sigma\,dt$ separate:

h(t)=\frac1{\sigma\sqrt{2\pi}}e^{-\frac12t^2}

Evaluate the two integrals separately:

First integral $\int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt = 0$ because $g(t)=t e^{-\frac{1}{2}t^2}$ is odd. Thus $g(-t) = -g(t)$ , and the signed areas over the symmetric interval $(-\infty, \infty)$ cancel.
Second integral $\int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt = \sqrt{2\pi}$ is the Gaussian integral. After multiplication by the normalizing factor $1/\sqrt{2\pi}$ , it gives total probability $1$ for the standard normal density.

The calculation gives:

E(X) = \frac{\sigma}{\sqrt{2\pi}} \cdot 0 + \frac{\mu}{\sqrt{2\pi}} \cdot \sqrt{2\pi}

The expected value of a normal random variable equals its mean parameter.

For every normal distribution $X \sim N(\mu, \sigma^2)$ , we have $E(X) = \mu$ . This means we don't need to perform integration every time we calculate the expected value of a normal distribution. We simply use the parameter value $\mu$ .

Practical Interpretation

For a normal distribution, the expected value and mean parameter are related as follows.

Example:

If the heights of a group of students are normally distributed with mean $165 \text{ cm}$ , then the expected height of a randomly selected student is $165 \text{ cm}$ .

Expected Values for Exam Scores and Birth Weights

Example 1

Suppose the math exam scores in a class are normally distributed with mean $\mu = 75$ and standard deviation $\sigma = 10$ . What is the expected value of a student's exam score?

Solution:

Since normal distribution has the property $E(X) = \mu$ , the expected value of the exam score is:

E(X) = 75

The expected value of a student's exam score is $75$ .

Example 2

Solution:

Given $\mu = 3{,}200 \text{ grams}$ and $\sigma = 400 \text{ grams}$ .

The expected weight is the mean parameter:

E(X) = \mu = 3{,}200 \text{ grams}

Exercises

Travel time from home to school is modeled by a normal distribution with mean $25 \text{ minutes}$ and standard deviation $5 \text{ minutes}$ . Determine the expected travel time.
Daily air temperature in Jakarta during June is modeled by a normal distribution with mean $28^\circ\text{C}$ and standard deviation $3^\circ\text{C}$ . What is the expected temperature on a randomly selected day?
Physics exam scores of grade $12$ students are modeled by a normal distribution with mean $78$ and standard deviation $12$ . What is the expected score of a randomly selected student?

Answer Key

Solution to Problem 1:

Given: $\mu = 25 \text{ minutes}$ , $\sigma = 5 \text{ minutes}$

Since normal distribution has the property $E(X) = \mu$ , then:

$E(X) = 25 \text{ minutes}$

Answer: The expected value of travel time is $25 \text{ minutes}$ .
Solution to Problem 2:

Given: $\mu = 28^\circ\text{C}$ , $\sigma = 3^\circ\text{C}$

Using the basic property of normal distribution:

$E(X) = \mu = 28^\circ\text{C}$

Answer: The expected value of air temperature is $28^\circ\text{C}$ .
Solution to Problem 3:

Given: $\mu = 78$ , $\sigma = 12$

A normally distributed variable satisfies $E(X)=\mu$ . Therefore:

$E(X) = \mu = 78$

Answer: The expected exam score is $78$ .