**Standard Normal Distribution Formula – (Table of Contents)**

## What is the Standard Normal Distribution Formula?

Standard Normal Distribution is a type of probability distribution which is symmetric about the average or the mean, depicting that the data near the average or the mean are occurring more frequently when compared to the data which is far from the average or the mean. A score on the standard normal distribution can be termed as the “Z-score”.

Z – Score Formula is represented as below-

**Z – Score = ( X – µ ) / σ**

Where,

- X is a normal random variable
- µ is the average or the mean
- σ is the standard deviation

Then we need to derive probability from the above table.

### Explanation of the Standard Normal Distribution Formula

The standard normal distribution in order words referred to as the Z-distribution has the following properties:

- It has an average or says the mean of zero.
- It has a standard deviation which is equal to 1.

Using the standard normal table, we can find out the areas under the density curve. Z-score is sore on the standard normal distribution and should be interpreted as the number of standard deviations where the data point is below or above the average or the mean.

A negative Z-Score shall indicate a score which is below the mean or the average while A positive Z-Score shall indicate that the data point is above the mean or the average.

The standard normal distribution follows 68-95-99.70 Rule which is also called as the Empirical Rule, and as per that Sixty eight percent of the given data or the values shall fall within 1 standard deviation of the average or the mean, while ninety-five percent shall fall within 2 standard deviations, and finally the ninety-nine decimal seven percent of the value or the data shall fall within 3 standard deviations of the average or of the mean.

### Examples of Standard Normal Distribution Formula (with Excel Template)

Let’s see some simple to advanced examples of standard normal distribution equation to understand it better.

#### Example #1

**Consider mean given to you like 850, standard deviation as 100. You are required to calculate Standard Normal Distribution for a score above 940.**

**Solution:**

Use the following data for the calculation of standard normal distribution.

So, the calculation of z score^{ }can be done as follows-

Z – score = ( X – µ ) / σ

4.9 (927 ratings)

= (940 – 850) / 100

Z Score will be –

**Z Score = 0.90**

Now using the above table of the standard normal distribution, we have value for 0.90 as 0.8159 and we need to calculate the score above that which is P(Z >0.90).

We need the right path to the table. Hence, the probability would be 1 – 0.8159 which is equal to 0.1841.

Thus, only 18.41% of the scores lie above 940.

#### Example #2

**Sunita takes private tuition classes for mathematics subject and currently, she has around 100 students enrolled under her. After the 1 ^{st} test she took for her students she got followings average numbers scored by them and have ranked them percentile wise. **

**Solution:**

First, we to plot what are we targeting for which is the left side of the cure. P(Z<75).

Use the following data for the calculation of standard normal distribution.

For that, we need to calculate the mean and the standard deviation first.

The calculation of mean can be done as follows-

Mean = (98 + 40 + 55 + 77 + 76 + 80 + 85 + 82 + 65 + 77) / 10

Mean = 73.50

The calculation of standard deviation can be done as follows-

Standard deviation = √[∑(x – x) / (n-1)]

Standard deviation = 16.38

So, the calculation of z score^{ }can be done as follows-

Z – score= ( X – µ ) / σ

= (75 – 73.50) / 16.38

Z Score will be –

**Z Score = 0.09**

Now using the above table of a standard normal distribution, we have value for 0.09 as 0.5359 and that is the value for P (Z <0.09).

Hence 53.59% of the students scored below 75.

#### Example #3

**Vista limited is an electronic equipment showroom. It wants to analyze its consumer behavior. It has around 10,000 customers around the city. On average the customer spends 25,000 when it comes to its shop. However, the spending varies significantly as customers spend from 22,000 to 30,000 and the average of this variance around 10,000 customers that management of vista limited has come up with is around 500. **

**The management of Vista limited has approached you and they are interested to know what proportion of their customers spend more than 26,000? Assume that customers spend figures are normally distributed.**

**Solution:**

First, we to plot what are we targeting for which is the left side of the cure. P(Z>26000).

Use the following data for the calculation of standard normal distribution.

The calculation of z score^{ }can be done as follows-

Z – score= ( X – µ ) / σ

=(26000 – 25000) / 500

Z Score will be-

**Z Score = 2**

The calculation of standard normal distribution^{ }can be done as follows-

Standard normal distribution will be-

Now using the above table of the standard normal distribution, we have value for 2.00 which is 0.9772 and now we need to calculate for P(Z >2).

We need the right path to the table. Hence, the probability would be 1 – 0.9772 which is equal to 0.0228.

Hence 2.28% of the consumers spend above 26000.

### Relevance and Use

To make an informed and a proper decision, one needs to convert all of the scores to a similar scale. One needs to standardize those scores, converting all of them to the standard normal distribution using the Z score method, with a single standard deviation and a single average or the mean. Majorly this is used in the field of statistics and also in the field of finance that too by traders.

Many statistical theories have attempted to model the prices of the asset (in fields of finance) under the main assumption that they shall follow this kind of normal distribution. Price distributions mostly tend to have fatter tails, and, hence have kurtosis which is greater than 3 in real life scenarios. Such assets have been observed to have the price movements which are greater than 3 standard deviations beyond the average or the mean and more often than the expected assumption in a normal distribution.

### Recommended Articles

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