Standard Normal Z Table
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
\[Z = \frac{X - \mu}{\sigma}\]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:
\[z = 0.16\]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:
\[P ( Z < 0.16 ) = 0.5636\]
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
\[P ( Z > 0.16 ) = 1 - 0.5636\]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 $Z = 0.16$:
\[P ( 0 < Z < 0.16 ) = P ( Z < 0.16 ) - P ( Z < 0 )\]Substituting the corresponding values gives
\[P ( 0 < Z < 0.16 ) = 0.5636 - 0.5 = 0.0636\]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 $0.16$:
\[P ( - 0.16 < Z < 0.16 ) = 2 \times P ( 0 < Z < 0.16 ) = 0.1272\]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:
\[z = \frac{x - \mu}{\sigma}\]where $x$ represents the observed value of the random variable, which in this case corresponds to 6.5 hours. We have:
\[z = \frac{6.5 - 8.0}{1.2} = \frac{- 1.5}{1.2} = - 1.25\]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
\[P ( Z < - 1.25 ) = 0.1056\]