# Lab 4 - Normal distribution

## Class video

The class video is attached here so that you can watch my lecture again when you prepare the exams.

- If you have questions about my lecture, please use
**the comment section**at the bottom of this documents.

## Normal distribution

From the basic statistics course, we are already familiar with a bell-shape distribution called Normal distribution.

Here are some characteristics of all Normal curves;

```
- Symmetric, Unimodal
- It can be specified by its mean and standard deviation.
- Mean is related to center of the distribution.
- Sd controls the spread of the distribution.
- All Normal distributions obey the 68-95-99.7 rule.
```

Our textbook use the following notation for refering a specific normal distribution;

\[ \mathcal{N}(\mu, \sigma) \] where \(\mu\) represents the mean and \(\sigma\) represents the standard deviation.

Let us look at the impact of the mean and standard deviation on the shape of Normal distribtuions.

## The 68-95-99.7 rule

The rule is as follows;

```
- Approx. 68% of the obs. fall within one std. range from the mean.
- Approx. 95% of the obs. fall within two std. range from the mean.
- Approx. 99.7% of the obs. fall within three std. range from the mean.
```

## The standard Normal distrbituion

There is a very special Normal distribution among many Normal distribution. The standard Normal distribution is the Normal distribution with

\[ \mu = 0, \sigma = 1. \]

The capital alphabet \(Z\) is often used for a variable follows the standard Normal distribution. For this particular normal distribution, we will use the following notation:

\[ Z \sim \mathcal{N}\left(1, 0 \right) \]

### Find prob. of \(Z\) when the variable follows \(\mathcal{N}\left(1, 0 \right)\)

If we knows that the variable \(Z\) follows the standard Normal distribution, then we can find any probability (or proportion) using the standard Normal table.

#### Example (from our Lecture note)

Suppose we live in a particular Scandinavian city, where temperature is measured in Centigrade. Weather records kept for many years indicate that the temperature at 11:00 a.m. on Jan. 28 follows a standard normal distribution. What proportion of years we can expect the temperature at this time to be less than or equal to \(-1.5\)°C?

## Answer

Since the temperature follows the standard Normal distribution, I can use \(Z_{s. city}\) to represent the temperature of Scandinavian city. We can write the following:

\[ Z_{s. city} \sim \mathcal{N}(0, 1) \]

Also the proportion that we want to find can be written as,

\[ P(Z_{s. city} \le -1.5) \]

The above form corresponds to the table in p. 696 from the table pdf. From the first table, if we find \(-1.5\) from the row names and \(0.00\) from the coloum names, we can find the number \(0.0668\) which is our answer.

Thus, we have the following:

\[ P(Z_{s. city} \le -1.5) = 0.0668 \]

## Standardization: Convert Normal distrbituion to standard Normal distribution

We have learned how to find a probability of proportions with a variable which follows standard Normal distribution in the previous section. Now, we are going to learn the method of convert a varible following an arbitrary Normal distribution to the one follwoing the standard Normal distribution.

\[ X \sim \mathcal{N}(\mu, \sigma) \overset{convert}{\longrightarrow} Z \sim \mathcal{N}(0, 1) \]

All normal distributions would be the same if we measured in units of size \(\sigma\) around the mean \(\mu\) as center! If \(x\) is an observation from a distribution that has mean \(\mu\) and standard deviation \(\sigma\), the standardized value of \(x\) is

\[
z = \frac{x - \mu}{\sigma}
\]
Standardized values are often called *z-scores*.

#### Example: Homework Scores

Professor uploads the distribution of Homework 1 score in ICON as above and anounced that the Homework scores follows Normal distribution with \(\mu =10\), \(\sigma = 2\).

Q1. You checked your score out and found that your score was 12. What is the *z-score* of your homework score?

## Answer

\[ z_{score} = \frac{x_{score} - 10}{2} = \frac{12 - 10}{2} = 1. \]

Q2. We have 75 people registered in this course. How many people got the homework scores which are below yours? (You can approximate it using the proportion method)

## Answer

Let us denote \(X\) by the homework score of students in this class, and we know that it follows Normal distribution according to the professor: \[ X_{score} \sim \mathcal{N}(10, 2) \]

First, we need to figure out what is the proportion of people whose homework score is below 12. We can write it down as follows using statistical notation:

\[ P(X_{score} \le 12) = ? \] Now, we cannot calculate this proportion directly, since \(X_{score}\) does not follow standard Normal distribution rather it follows \(\mathcal{N}(10, 2)\). How can we conver this \(X\) to \(Z\)? Standardization formula!

According to standardization formula, we know the following:

\[ Z_{score} = \frac{X_{score} - 12}{2} \sim \mathcal{N}(0, 1) \]

If you find the *z-score* in the standard Normal table, you will know that
\[
P(X_{score} \le 12) \overset{stdardization}{=} P(Z_{score} \le 1) \overset{z-table}{=} 0.8413
\]

Thus, we can conclude that there are 84.13% of people in this class have the homework score below 12, which is your score. So our final answer will be

\[ 75 \times 0.8413 \approx 64 \]

About 64 students.

**Exercise.** How many people got the homework scores which are between 10 and 12?

##
Answer

\[ 75 \times 0.3413 \approx 26 \]

About 26 students. Why?**Exercise (Hard but MUST be in the exam).** The professor told you that you have made a top 8% of the class room based on the homework score. Can you guess your homework score?

##
Answer

- Step 1. Find the \(z\) score which corresponds to 0.92. (why?) \(z\) score is 1.41.
- Step 2. Convert \(z\) score to the homework distribution scale. \[ \begin{align*} your\underline{ }score & =\mu+\sigma\times z\\ & =10+2\times1.41\\ & =12.82 \end{align*} \] Your score is around 12.82.