# 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?

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?

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

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)

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

$75 \times 0.8413 \approx 64$

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

$75 \times 0.3413 \approx 26$
• 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.