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Understand expected value in normal distributions, why E(X)=μ, and how mean outcomes appear in statistics problems.

---

## 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 $$\mu$$.

Visible text: > For a normal distribution , its expected value is always equal to the mean parameter .

Mathematically, we can write:

```math
E(X) = \mu
```

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

Visible text: where is a random variable that follows normal distribution and 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:

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

For normal distribution, the probability density function is:

```math
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:

```math
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 $$dx = \sigma \, dt$$.

Visible text: Now, let's perform the substitution . Then and .

Component: MathContainer
Children:

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

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

```math
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 $$\sqrt{2\pi}$$.

Visible text: - **First integral** because the function is an odd function. This means , so when integrated over the symmetric interval , the results cancel each other out and equal zero.

- **Second integral** because this is the fundamental Gaussian integral. This integral represents the total area under the standard normal distribution curve, which always equals .

Therefore:

Component: MathContainer
Children:

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

```math
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 $$\mu$$.

Visible text: > For every normal distribution , we have . 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 .

## 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 $$\mu$$.

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

**Simple example:**

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

Visible text: If the height of students in your school is normally distributed with mean , then the expected value of the height of a randomly selected student is .

## Application Examples

**Example** $$1$$

Visible text: **Example**

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?

Visible text: Suppose the math exam scores in a class are normally distributed with mean and standard deviation . 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:

Visible text: Since normal distribution has the property , the expected value of the exam score is:

```math
E(X) = 75
```

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

Visible text: Therefore, the expected value of a student's exam score is .

**Example** $$2$$

Visible text: **Example**

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

Visible text: The weight of newborn babies at a hospital is normally distributed with mean and standard deviation . Determine the expected value of the weight of a baby to be born!

**Solution:**

Given $$\mu = 3.200 \text{ grams}$$ and $$\sigma = 400 \text{ grams}$$.

Visible text: Given and .

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

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

## Exercises

1. Travel time from home to school is normally distributed with mean $$25 \text{ minutes}$$ and standard deviation $$5 \text{ minutes}$$. Determine the expected value of travel time!

2. Daily air temperature in Jakarta during June is normally distributed with mean $$28^\circ\text{C}$$ and standard deviation $$3^\circ\text{C}$$. What is the expected value of air temperature on a randomly selected day?

3. 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?

Visible text: 1. Travel time from home to school is normally distributed with mean and standard deviation . Determine the expected value of travel time!

2. Daily air temperature in Jakarta during June is normally distributed with mean and standard deviation . What is the expected value of air temperature on a randomly selected day?

3. Physics exam scores of grade students are normally distributed with mean and standard deviation . If a student takes the exam, what is the expected value they will obtain?

### Answer Key

1. **Solution to Problem** $$1$$:

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

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

   
   
   ```math
   E(X) = 25 \text{ minutes}
   ```

   **Answer:** The expected value of travel time is $$25 \text{ minutes}$$.

2. **Solution to Problem** $$2$$:

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

   Using the basic property of normal distribution:

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

   **Answer:** The expected value of air temperature is $$28^\circ\text{C}$$.

3. **Solution to Problem** $$3$$:

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

   Based on the fundamental property of normal distribution:

   
   
   ```math
   E(X) = \mu = 78
   ```

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

Visible text: 1. **Solution to Problem** :

 Given: , 

 Since normal distribution has the property , then:

 
 

 **Answer:** The expected value of travel time is .

2. **Solution to Problem** :

 Given: , 

 Using the basic property of normal distribution:

 
 

 **Answer:** The expected value of air temperature is .

3. **Solution to Problem** :

 Given: , 

 Based on the fundamental property of normal distribution:

 
 

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