Standard Normal Z Table

Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/standard-normal-z-table/ Fetched from algebrica.org post 18966; source modified 2026-03-14T21:54:56.

From the normal to the standard normal distribution

A generic normal distribution $\mathcal{N}(x; \mu, \sigma)$ can always be transformed into its standardized form $N(x; 0, 1)$ by introducing the standard variable $Z$, defined as

This transformation, known as standardization, expresses each value of the continuous random variable $X$ in terms of the number of standard deviations it lies away from the mean $\mu$. As a result, the new variable $Z$ follows a standard normal distribution with mean $0$ and standard deviation $1$.

The process of standardization is particularly useful because it places different normal distributions on a common scale. Once the variable $X$ has been transformed into the standard variable $Z$, probabilities and critical values can be derived directly from the standard normal distribution $N(x; 0, 1)$.

This allows statistical problems to be solved using a single universal reference table (Z table), simplifying calculations and comparisons across different contexts.

Standard Z Table

The standard Z table lists, for each cell, the cumulative probability to the left of a given $Z$ value. In other words, it shows the probability that a standard normal variable takes on a value less than or equal to a specified Z-score. These values correspond to the area under the standard normal curve from negative infinity up to the chosen point on the horizontal axis, represented by the value of $z$.

The complete Z table is extensive, as it contains values for a wide range of Z-scores. For brevity, only a portion of the table is shown below as an illustrative extract.

z

.00

.01

.02

.03

.04

.05

.06

.07

.08

…

0.0

.5000

.5040

.5080

.5120

.5160

.5199

.5239

.5279

.5319

…

0.1

.5398

.5438

.5478

.5517

.5557

.5596

.5636

.5675

.5714

…

0.2

.5793

.5832

.5871

.5910

.5948

.5987

.6026

.6064

.6103

…

0.3

.6179

.6217

.6255

.6293

.6331

.6368

.6406

.6443

.6480

…

0.4

.6554

.6591

.6628

.6664

.6700

.6736

.6772

.6808

.6844

…

0.5

.6915

.6950

.6985

.7019

.7054

.7088

.7123

.7157

.7190

…

0.6

.7257

.7291

.7324

.7357

.7389

.7422

.7454

.7486

.7517

…

0.7

.7580

.7611

.7642

.7673

.7704

.7734

.7764

.7794

.7823

…

0.8

.7881

.7910

.7939

.7967

.7995

.8023

.8051

.8078

.8106

…

0.9

.8159

.8186

.8212

.8238

.8264

.8289

.8315

.8340

.8365

…

1.0

.8413

.8438

.8461

.8485

.8508

.8531

.8554

.8577

.8599

…

…

…

…

…

…

…

…

…

…

…

…

Several online resources, both static and interactive, allow users to compute cumulative probabilities for different values of $z$.

Using the Z table is straightforward.

• The row corresponding to the $Z$ value contains the integer part and the first decimal place
• The column represents the remaining decimal places starting from the second one.
• The cell at the intersection of the selected row and column gives the cumulative probability up to the specified value of $z$.

Example 1

Let us consider a simple example to illustrate how the Z table is used to determine the cumulative probability corresponding to a given value of the standardized variable $Z$. Let us consider the case of a standardized variable $Z$ such that:

To find the cumulative probability to the left of the standardized variable $z$, we locate the value 0.1 in the row and 0.06 in the column of the Z table. The intersection of these two entries gives the corresponding probability value, which in this case is $0.5636$.

z

.00

.01

.02

.03

.04

.05

.06

.07

.08

…

0.0

.5000

.5040

.5080

.5120

.5160

.5199

.5239

.5279

.5319

…

0.1

.5398

.5438

.5478

.5517

.5557

.5596

.5636

.5675

.5714

…

…

…

…

…

…

…

…

…

…

…

…

This means that approximately 56.36% of the observations in a standard normal distribution fall below this value of $z$. We can therefore state that the probability of the standardized variable $Z$ being less than $z = 0.16$ is equal to $0.5636$, that is:

Since the total area under the standard normal curve equals $1$, and the distribution is symmetric with respect to its mean, we can immediately deduce that

Moreover, the probability that $Z$ lies between $0$ and $0.16$ is obtained by subtracting the cumulative probability up to $Z = 0$ from that up to $Z = 0.16$:

Substituting the corresponding values gives

Finally, due to the symmetry of the normal distribution about its mean, the probability that $Z$ lies within the interval $-0.16 < Z < 0.16$ is twice the probability of being between $0$ and $0.16$:

Example 2

Suppose that the average lifetime of a rechargeable battery is $\mu = 8.0$ hours, with a standard deviation of $\sigma = 1.2$ hours, and that battery life follows a normal distribution. We want to calculate the probability that a randomly selected battery lasts less than $6.5$ hours.

To find the probability $P(X < 6.5)$, we need to determine the area under the normal curve to the left of $6.5$. For this purpose, we apply the standardization formula:

where $x$ represents the observed value of the random variable, which in this case corresponds to 6.5 hours. We have:

We can then use the Z table to find $P(Z < -1.25)$, which gives the cumulative probability associated with this value of $z$.

z

.00

.01

.02

.03

.04

.05

.06

.07

.08

…

-1.0

.1587

.1562

.1539

.1515

.1492

.1469

.1446

.1423

.1401

…

-1.2

.1151

.1131

.1112

.1093

.1075

.1056

.1038

.1020

.1003

…

-1.3

.0968

.0951

.0934

.0918

.0901

.0885

.0869

.0853

.0838

…

…

…

…

…

…

…

…

…

…

…

…

From the intersection of the selected row and column, we obtain a probability value equal to $0.1056$. The figure below illustrates where this probability is located under the standard normal curve, corresponding to the cumulative area to the left of $z = -1.25$.

From the standard normal table, we obtain

This means that approximately 10.56% of all batteries are expected to last less than 6.5 hours.