# Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one.

The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

z = (X - μ) / σ

where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X.

## Standard Normal Distribution Table

A standard normal distribution table shows a cumulative probability associated with a particular z-score. Table rows show the whole number and tenths place of the z-score. Table columns show the hundredths place. The cumulative probability (often from minus infinity to the z-score) appears in the cell of the table.

For example, a section of the standard normal table is reproduced below. To find the cumulative probability of a z-score equal to -1.31, cross-reference the row of the table containing -1.3 with the column containing 0.01. The table shows that the probability that a standard normal random variable will be less than -1.31 is 0.0951; that is, P(Z < -1.31) = 0.0951.

 z 0.00 0.01 0.02 0.03 0.04 0.05 -3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 ... ... ... ... ... ... ... -1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 -1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 -1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 ... ... ... ... ... ... ... 3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989

Of course, you may not be interested in the probability that a standard normal random variable falls between minus infinity and a given value. You may want to know the probability that it lies between a given value and plus infinity. Or you may want to know the probability that a standard normal random variable lies between two given values. These probabilities are easy to compute from a normal distribution table. Here's how.

• Find P(Z > a). The probability that a standard normal random variable (z) is greater than a given value (a) is easy to find. The table shows the P(Z < a). The P(Z > a) = 1 - P(Z < a).

Suppose, for example, that we want to know the probability that a z-score will be greater than 3.00. From the table (see above), we find that P(Z < 3.00) = 0.9987. Therefore, P(Z > 3.00) = 1 - P(Z < 3.00) = 1 - 0.9987 = 0.0013.

• Find P(a < Z < b). The probability that a standard normal random variables lies between two values is also easy to find. The P(a < Z < b) = P(Z < b) - P(Z < a).

For example, suppose we want to know the probability that a z-score will be greater than -1.40 and less than -1.20. From the table (see above), we find that P(Z < -1.20) = 0.1151; and P(Z < -1.40) = 0.0808. Therefore, P(-1.40 < Z < -1.20) = P(Z < -1.20) - P(Z < -1.40) = 0.1151 - 0.0808 = 0.0343.

In school or on the Advanced Placement Statistics Exam, you may be called upon to use or interpret standard normal distribution tables. Standard normal tables are commonly found in appendices of most statistics texts.

## The Normal Distribution as a Model for Real-World Events

Often, phenomena in the real world follow a normal (or near-normal) distribution. This allows researchers to use the normal distribution as a model for assessing probabilities associated with real-world phenomena. Typically, the analysis involves two steps.

• Transform raw data. Usually, the raw data are not in the form of z-scores. They need to be transformed into z-scores, using the transformation equation presented earlier: z = (X - μ) / σ.

• Find probability. Once the data have been transformed into z-scores, you can use standard normal distribution tables, online calculators (e.g., Stat Trek's free normal distribution calculator), or handheld graphing calculators to find probabilities associated with the z-scores.

The problem in the next section demonstrates the use of the normal distribution as a tool to model real-world events.

## Test Your Understanding

Problem 1

Molly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)

(A) 0.10
(B) 0.18
(C) 0.50
(D) 0.82
(E) 0.90

Solution

The correct answer is B. As part of the solution to this problem, we assume that test scores are normally distributed. In this way, we use the normal distribution to model the distribution of test scores in the real world. Given an assumption of normality, the solution involves three steps.

• First, we transform Molly's test score into a z-score, using the z-score transformation equation.

z = (X - μ) / σ = (940 - 850) / 100 = 0.90

• Then, using an online calculator (e.g., Stat Trek's free normal distribution calculator), a handheld graphing calculator, or the standard normal distribution table, we find the cumulative probability associated with the z-score. In this case, we find P(Z < 0.90) = 0.8159.

• Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.

Thus, we estimate that 18.41 percent of the students tested had a higher score than Molly.