Expected Value of Normal Distribution: Data Analysis and Probability - Mathematics (Grade 12)

Basic Concept of Expected Value

Have you ever wondered, if we repeatedly take samples from a normal distribution, what value appears most frequently? Or in other words, what is the central value that we expect from that distribution?

This is what we call expected value. In the context of normal distribution, expected value has a very interesting and simple property.

For a normal distribution $n(x; \mu, \sigma)$ , its expected value is always equal to the mean parameter $\mu$ .

Mathematically, we can write:

E(X) = \mu

where $X$ is a random variable that follows normal distribution and $\mu$ is the mean parameter of that distribution.

Why is this so? Let's prove it mathematically.

Mathematical Proof

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

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

For normal distribution, 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

Now, let's perform the substitution $t = \frac{x-\mu}{\sigma}$ . Then $x = \sigma t + \mu$ and $dx = \sigma \, dt$ .

E(X) = \int_{-\infty}^{\infty} (\sigma t + \mu) \cdot \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}t^2} \cdot \sigma \, dt

E(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (\sigma t + \mu) e^{-\frac{1}{2}t^2} \, dt

E(X) = \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt + \frac{\mu}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt

Note that:

First integral $\int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt = 0$ because the function $t e^{-\frac{1}{2}t^2}$ is an odd function. This means $f(-t) = -f(t)$ , so when integrated over the symmetric interval $[-\infty, \infty]$ , the results cancel each other out and equal zero.
Second integral $\int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt = \sqrt{2\pi}$ because this is the fundamental Gaussian integral. This integral represents the total area under the standard normal distribution curve, which always equals $\sqrt{2\pi}$ .

Therefore:

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

E(X) = 0 + \mu = \mu

It is proven that the expected value of normal distribution 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

The proof result above provides a very important understanding:

If we have a normal distribution with mean $\mu$ and standard deviation $\sigma$ , then the expected value of that random variable is exactly $\mu$ . This means if we take samples in very large numbers and calculate their average, the result will approach the value $\mu$ .

Simple example:

If the height of students in your school is normally distributed with mean $165$ cm, then the expected value of the height of a randomly selected student is $165$ cm.

Application Examples

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

Therefore, the expected value of a student's exam score is $75$ .

Example 2

The weight of newborn babies at a hospital is normally distributed with mean $3.200$ grams and standard deviation $400$ grams. Determine the expected value of the weight of a baby to be born!

Solution:

Given $\mu = 3.200$ grams and $\sigma = 400$ grams.

The expected value of the weight of a baby to be born is:

E(X) = \mu = 3.200 \text{ grams}

Exercises

Travel time from home to school is normally distributed with mean $25$ minutes and standard deviation $5$ minutes. Determine the expected value of travel time!
Daily air temperature in Jakarta during June is normally distributed with mean $28°C$ and standard deviation $3°C$ . What is the expected value of air temperature on a randomly selected day?
Physics exam scores of grade 12 students are normally distributed with mean $78$ and standard deviation $12$ . If a student takes the exam, what is the expected value they will obtain?

Answer Key

Solution to Problem 1:

Given: $\mu = 25$ minutes, $\sigma = 5$ minutes

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

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

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

Given: $\mu = 28°C$ , $\sigma = 3°C$

Using the basic property of normal distribution:

$E(X) = \mu = 28°C$

Answer: The expected value of air temperature is $28°C$ .
Solution to Problem 3:

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

Based on the fundamental property of normal distribution:

$E(X) = \mu = 78$

Answer: The expected value that the student will obtain is $78$ .

Basic Concept of Expected Value

This is what we call expected value. In the context of normal distribution, expected value has a very interesting and simple property.

For a normal distribution $n(x; \mu, \sigma)$ , its expected value is always equal to the mean parameter $\mu$ .

Mathematically, we can write:

E(X) = \mu

where $X$ is a random variable that follows normal distribution and $\mu$ is the mean parameter of that distribution.

Why is this so? Let's prove it mathematically.

Mathematical Proof

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

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

For normal distribution, 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

Now, let's perform the substitution $t = \frac{x-\mu}{\sigma}$ . Then $x = \sigma t + \mu$ and $dx = \sigma \, dt$ .

E(X) = \int_{-\infty}^{\infty} (\sigma t + \mu) \cdot \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}t^2} \cdot \sigma \, dt

E(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} (\sigma t + \mu) e^{-\frac{1}{2}t^2} \, dt

E(X) = \frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt + \frac{\mu}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt

Note that:

First integral $\int_{-\infty}^{\infty} t e^{-\frac{1}{2}t^2} \, dt = 0$ because the function $t e^{-\frac{1}{2}t^2}$ is an odd function. This means $f(-t) = -f(t)$ , so when integrated over the symmetric interval $[-\infty, \infty]$ , the results cancel each other out and equal zero.
Second integral $\int_{-\infty}^{\infty} e^{-\frac{1}{2}t^2} \, dt = \sqrt{2\pi}$ because this is the fundamental Gaussian integral. This integral represents the total area under the standard normal distribution curve, which always equals $\sqrt{2\pi}$ .

Therefore:

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

E(X) = 0 + \mu = \mu

It is proven that the expected value of normal distribution 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

The proof result above provides a very important understanding:

Simple example:

If the height of students in your school is normally distributed with mean $165$ cm, then the expected value of the height of a randomly selected student is $165$ cm.

Application Examples

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

Therefore, the expected value of a student's exam score is $75$ .

Example 2

The weight of newborn babies at a hospital is normally distributed with mean $3.200$ grams and standard deviation $400$ grams. Determine the expected value of the weight of a baby to be born!

Solution:

Given $\mu = 3.200$ grams and $\sigma = 400$ grams.

The expected value of the weight of a baby to be born is:

E(X) = \mu = 3.200 \text{ grams}

Exercises

Travel time from home to school is normally distributed with mean $25$ minutes and standard deviation $5$ minutes. Determine the expected value of travel time!
Daily air temperature in Jakarta during June is normally distributed with mean $28°C$ and standard deviation $3°C$ . What is the expected value of air temperature on a randomly selected day?
Physics exam scores of grade 12 students are normally distributed with mean $78$ and standard deviation $12$ . If a student takes the exam, what is the expected value they will obtain?

Answer Key

Solution to Problem 1:

Given: $\mu = 25$ minutes, $\sigma = 5$ minutes

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

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

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

Given: $\mu = 28°C$ , $\sigma = 3°C$

Using the basic property of normal distribution:

$E(X) = \mu = 28°C$

Answer: The expected value of air temperature is $28°C$ .
Solution to Problem 3:

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

Based on the fundamental property of normal distribution:

$E(X) = \mu = 78$

Answer: The expected value that the student will obtain is $78$ .

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