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 mean of X, and σ is the standard deviation of
Standard Normal Distribution Table
A standard normal distribution table shows a
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
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.
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
The Normal Distribution as a Model for Measurements
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
- 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),
to find probabilities associated with the z-scores.
The problem in the next section demonstrates the use of the normal
distribution as a model for measurement.
Test Your Understanding
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.)
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
as a model for measurement. Given an assumption of normality, the
solution involves three steps.
Thus, we estimate that 18.41 percent of the students tested had a
higher score than Molly.