Given the Mean and Standard Deviation, How Do You Find a Probability?

The mean $\mu$ and standard deviation $\sigma$ alone do not determine a probability: you also need the shape of the distribution. In most textbook problems that shape is normal (often stated as “assume the data are normally distributed”), and then everything follows from one step: the z-score.

Step 1: Convert to a z-score

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

The z-score counts how many standard deviations $x$ is above (positive) or below (negative) the mean. If $X \sim N(\mu, \sigma^2)$, then $Z = (X - \mu)/\sigma$ follows the standard normal distribution $N(0, 1)$, whose cumulative probabilities $\Phi(z) = P(Z < z)$ are tabulated.

Step 2: Use the right formula

You want	Formula
$P(X < b)$	$\Phi(z_b)$
$P(X > a)$	$1 - \Phi(z_a)$
$P(a < X < b)$	$\Phi(z_b) - \Phi(z_a)$

For a continuous distribution, $<$ and $\le$ give the same probability.

Worked example

Exam scores are normally distributed with $\mu = 70$ and $\sigma = 10$.

(a) What proportion of students score above 85?

$$z = \frac{85 - 70}{10} = 1.5, \qquad P(X > 85) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$$

About 6.7% of students.

(b) What proportion score between 60 and 80?

$$z_a = \frac{60-70}{10} = -1, \quad z_b = \frac{80-70}{10} = 1$$ $$P(60 < X < 80) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6827$$

This is the familiar “68% within one standard deviation” rule.

(c) What proportion score below 62?

$$z = \frac{62 - 70}{10} = -0.8, \qquad P(X < 62) = \Phi(-0.8) = 0.2119$$

If your table only lists positive $z$, use symmetry: $\Phi(-z) = 1 - \Phi(z) = 1 - 0.7881$.

Going backwards: from a probability to a value

To find the score that only the top 10% exceed, first find the $z$ with $\Phi(z) = 0.90$. From the table, $z \approx 1.2816$. Then undo the standardisation:

$$x = \mu + z\sigma = 70 + 1.2816 \times 10 \approx 82.8$$

Probabilities for a sample mean

If the question is about the average of $n$ observations, the standard deviation shrinks to the standard error $\sigma/\sqrt{n}$:

$$z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}$$

For example, the probability that 25 students average above 73 is $P\!\left(Z > \frac{73-70}{10/5}\right) = P(Z > 1.5) = 0.0668$. By the central limit theorem this works approximately even when individual scores aren’t normal, as long as $n$ is reasonably large.

What if the distribution isn’t normal?

With only $\mu$ and $\sigma$ and no shape assumption, you can’t get an exact probability. Chebyshev’s inequality still gives a bound that holds for any distribution:

$$P(|X - \mu| \ge k\sigma) \le \frac{1}{k^2}$$

For $k = 2$: at most 25% of values lie more than two standard deviations from the mean, whatever the distribution. That is much weaker than the normal distribution’s 4.6%, which shows how much the normality assumption is doing.

Common mistakes

• Dividing by the variance instead of the standard deviation. If you’re given $\sigma^2 = 100$, use $\sigma = 10$.
• Forgetting $1 - \Phi$ for “greater than” questions. Tables give the area to the left.
• Applying normal methods to clearly skewed data, such as waiting times or incomes, without justification.
• Using $\sigma$ instead of $\sigma/\sqrt{n}$ for questions about a sample mean.