What if the calculated z-score is not on the standard table?

If the z-score exceeds the table limits (typically $|z| > 3.49$), the cumulative probability is treated as 0 for negative values and 1 for positive values. For intermediate values not explicitly listed, use linear interpolation between the two nearest entries.

Does it matter if the probability inequality is strict ($<$) or inclusive ($\le$)?

No. Because the normal distribution is continuous, the probability of any exact single value is exactly zero. Therefore, $P(X < x)$ is mathematically identical to $P(X \le x)$.

How do you handle a sample mean instead of a single observation?

If evaluating a probability for a sample mean $\bar{X}$ of size $n$, replace the population standard deviation $\sigma$ with the standard error of the mean $\frac{\sigma}{\sqrt{n}}$ in the denominator of the z-score formula.

How to find a probability with the normal distribution

Q: Does it matter if the probability inequality is strict ($<$) or inclusive ($\le$)?

No. Because the normal distribution is continuous, the probability of any exact single value is exactly zero. Therefore, $P(X < x)$ is mathematically identical to $P(X \le x)$.

Q: How do you handle a sample mean instead of a single observation?

If evaluating a probability for a sample mean $\bar{X}$ of size $n$, replace the population standard deviation $\sigma$ with the standard error of the mean $\frac{\sigma}{\sqrt{n}}$ in the denominator of the z-score formula.

To find a probability with the normal distribution, convert the variable $X$ to a standard score $Z$ and evaluate the cumulative distribution function. This method applies whenever a continuous random variable is known or assumed to follow a normal distribution $N(\mu, \sigma^2)$ .

The setup

Identify the population mean $\mu$ , the population standard deviation $\sigma$ , and the target boundary values $x$ . Define the required probability bound, choosing among a left-tail $P(X < x)$ , a right-tail $P(X > x)$ , or an interval $P(x_1 < X < x_2)$ .

The steps

Calculate the z-score for each boundary value using $z = \frac{x - \mu}{\sigma}$ .
Look up the calculated z-score in a standard normal (Z) table or use a cumulative distribution function (CDF) calculator to find the left-tail probability $P(Z < z)$ .
Adjust the cumulative area for your specific bound: use $1 - P(Z < z)$ for right-tail probabilities, or $P(Z < z_2) - P(Z < z_1)$ for intervals.

Checking the result

Verify that the final probability is strictly between 0 and 1. Confirm that target values of $x$ close to the mean $\mu$ yield cumulative left-tail probabilities near 0.5, and values far above the mean yield cumulative probabilities near 1.

Common errors

A frequent error is using the variance $\sigma^2$ instead of the standard deviation $\sigma$ in the denominator of the z-score formula. Another standard mistake is failing to subtract the table value from 1 when evaluating a right-tail ('greater than') probability.

Worked example

The heights of a population are normally distributed with $\mu = 170$ cm and $\sigma = 10$ cm. Find the probability that a randomly selected person is between 165 cm and 180 cm tall.

Let $X \sim N(170, 10^2)$ . We need to find $P(165 < X < 180)$ .

Step 1: Calculate z-scores for both boundaries. $z_1 = \frac{165 - 170}{10} = -0.5$ $z_2 = \frac{180 - 170}{10} = 1.0$

Step 2: Look up cumulative probabilities in the Z-table. $P(Z < -0.5) = 0.3085$ $P(Z < 1.0) = 0.8413$

Step 3: Subtract the smaller area from the larger area to find the interval probability. $P(-0.5 < Z < 1.0) = 0.8413 - 0.3085 = 0.5328$

The probability is 0.5328.

FAQ

Run your own problem

Answer form

Drag an image or PDF of the problem here, or click to upload

Ctrl + V to paste

References: OpenStax Introductory Statistics, Chapter 6: The Normal Distribution · Khan Academy, Normal Distribution Unit