Standard Normal Z Table

From the normal to the standard normal distribution

1.If $X$ has a normal distribution $\mathcal{N}(\mu, \sigma^{2}),$ its standardized variable $Z$ is defined by:

$$Z = \frac{X - \mu}{\sigma}$$

2.This transformation is called standardization. For each value $x$ of the continuous random variable $X,$ the corresponding value $z$ is the difference between $x$ and the mean $\mu,$ measured in standard deviations. The variable $Z$ has the standard normal distribution $\mathcal{N}(0, 1),$ with mean $0$ and standard deviation $1.$

3.After standardization, every normal variable has the same distribution. Thus probabilities and critical values for any normal distribution can be calculated from $\mathcal{N}(0, 1)$ using one reference table, the Z table, instead of evaluating a separate integral for each pair $(\mu, \sigma).$

Standard Z table

4.The standard Z table lists, for each value of $z,$ the cumulative probability to its left. This probability is the cumulative distribution function of the standard normal variable, written $\Phi(z)$:

$$\Phi(z) = P(Z \le z)$$

5.The value $\Phi(z)$ is the area under the standard normal curve from $-\infty$ up to $z.$

6.Since the standard normal distribution is continuous, every single value has probability zero, so $P(Z = z) = 0.$ Including or excluding the endpoint $z$ therefore does not change the cumulative probability:

$$P(Z < z) = P(Z \le z)$$

7.

The normal distribution page defines $\Phi$ as an improper integral of the standard normal density and derives its properties. Here its values are read from the table.

8.The full table covers a wide range of $z$ values. The extract below shows a portion of it, with probabilities rounded to four decimal places.

9.

An interactive Z table computes cumulative probabilities for a given value of $z.$

10.The table is read by splitting $z$ into two parts.

Example 1

12.For the standard normal variable $Z$ at $z = 0.16,$ the cumulative probability to its left is found in row $0.1$ and column $0.06.$ The entry at their intersection is $0.5636,$ so $\Phi(0.16) \approx 0.5636.$

13.Thus about $56.36\%$ of values from the standard normal distribution are less than $0.16$:

$$P(Z < 0.16) \approx 0.5636$$

14.The total area under the curve equals $1,$ so the complementary probability is:

$$P(Z > 0.16) \approx 1 - 0.5636 = 0.4364$$

15.The probability that $Z$ lies between $0$ and $0.16$ is the difference between the cumulative values at $0.16$ and $0$:

$$P(0 < Z < 0.16) = \Phi(0.16) - \Phi(0) \approx 0.5636 - 0.5 = 0.0636$$

16.By symmetry about the mean, the interval $-0.16 < Z < 0.16$ has twice this probability:

$$P(-0.16 < Z < 0.16) = 2P(0 < Z < 0.16) \approx 0.1272$$

Combining table values

17.A single entry gives $\Phi(z) = P(Z \le z).$ Other probabilities follow from it by complementation and difference.

19.A table that lists only $z \ge 0$ can also be used for negative arguments through the symmetry of the curve about $0,$ where $\Phi(-z) = 1 - \Phi(z).$ The normal distribution page derives this identity.

20.For an interval with one negative endpoint, we combine both rules. For $a = -1.37$ and $b = 2.01,$ the interval probability is:

$$P(-1.37 < Z \le 2.01) = \Phi(2.01) - \Phi(-1.37) = \Phi(2.01) + \Phi(1.37) - 1$$

21.The four-decimal table entries are $\Phi(2.01) \approx 0.9778$ and $\Phi(1.37) \approx 0.9147,$ so:

$$P(-1.37 < Z \le 2.01) \approx 0.9778 + 0.9147 - 1 = 0.8925$$

22.

The normal approximation of the binomial distribution uses the same subtraction of cumulative values after the discrete bounds have been adjusted and standardized.

Example 2

23.The lifetime of a rechargeable battery is normally distributed with mean $\mu = 8.0$ hours and standard deviation $\sigma = 1.2$ hours. We find the probability that a battery lasts less than $6.5$ hours.

24.The probability $P(X < 6.5)$ is the area under the curve to the left of $6.5.$ We standardize the value:

$$z = \frac{x - \mu}{\sigma}$$

25.For $x = 6.5$ hours, the standardized value is:

$$z = \frac{6.5 - 8.0}{1.2} = \frac{-1.5}{1.2} = -1.25$$

26.The required probability is $P(Z < -1.25) = \Phi(-1.25).$

27.The entry in row $-1.2$ and column $0.05$ is $0.1056.$ The figure below shows this probability as the area under the standard normal curve to the left of $z = -1.25$:

$$P(Z < -1.25) \approx 0.1056$$

28.

The same value follows from symmetry for a table that lists only $z \ge 0,$ since $\Phi(-1.25) = 1 - \Phi(1.25) \approx 1 - 0.8944 = 0.1056.$

29.According to the model, about $10.56\%$ of the batteries last less than $6.5$ hours.

Reading the table in reverse

30.In some problems, the cumulative probability is given and $z$ is unknown. To determine $z,$ locate the probability in the body of the table and read the corresponding value from its row and column.

31.Suppose that the probability to the left of $z$ is $0.85.$ Then:

$$\Phi(z) = P(Z \le z) = 0.85$$

32.The entry closest to $0.85$ is $0.8508,$ in row $1.0$ and column $0.04,$ so:

$$z \approx 1.04$$

33.The value with a prescribed probability $\alpha$ to its left is the quantile $z_\alpha$ of the standard normal distribution. For example, $0.97$ lies between $\Phi(1.88) \approx 0.9699$ and $\Phi(1.89) \approx 0.9706,$ and is closer to $\Phi(1.88),$ so its quantile is approximately $1.88.$ The normal distribution page lists quantiles commonly used in interval estimation.