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

\[ 75 \times 0.8413 \approx 64 \]

About 64 students.

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


\[ 75 \times 0.3413 \approx 26 \]

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?

  • 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.