How to Compute Probabilities for a Normally Distributed Variable

A normally distributed random variable is determined by its mean and standard deviation. If

$$X \sim \mathcal{N}(\mu,\sigma^2),$$

then $\mu$ is the mean and $\sigma$ is the standard deviation. The main computation is to convert an observed value into a standard score, or $z$-score, and then use the standard normal distribution.

The standardization formula

For any value $x$, define

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

This tells you how many standard deviations $x$ is above or below the mean. The standardized variable

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

has the standard normal distribution, written $Z\sim\mathcal{N}(0,1)$.

Let $\Phi(z)$ denote the standard normal cumulative distribution function:

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

Because a normal distribution is continuous, $P(Z\le z)=P(Z<z)$; it makes no difference whether the inequality is strict.

The three basic probability calculations

Probability below a value

To calculate $P(X\le x)$, standardize $x$:

$$P(X\le x)=P\left(Z\le \frac{x-\mu}{\sigma}\right) =\Phi\left(\frac{x-\mu}{\sigma}\right).$$

Probability above a value

For an upper-tail probability, use the complement:

$$P(X\ge x)=1-\Phi\left(\frac{x-\mu}{\sigma}\right).$$

Probability between two values

For $a<b$,

$$P(a\le X\le b) =\Phi\left(\frac{b-\mu}{\sigma}\right) -\Phi\left(\frac{a-\mu}{\sigma}\right).$$

The subtraction is necessary because the CDF includes everything to the left of each boundary.

Worked example

Suppose the lifetime of a component is normally distributed with mean $\mu=100$ hours and standard deviation $\sigma=15$ hours. Find the probability that a randomly selected component lasts between 85 and 130 hours.

Write the model as

$$X\sim\mathcal{N}(100,15^2).$$

Standardize both endpoints:

$$z_1=\frac{85-100}{15}=-1,$$

and

$$z_2=\frac{130-100}{15}=2.$$

Therefore,

$$P(85\le X\le 130)=\Phi(2)-\Phi(-1).$$

Using standard normal values,

$$\Phi(2)\approx 0.97725, \qquad \Phi(-1)\approx 0.15866.$$

Thus,

$$P(85\le X\le 130) \approx 0.97725-0.15866 =0.81859.$$

So the probability is approximately $0.8186$, or 81.86%.

The same example also gives two useful tail probabilities. The probability of lasting less than 85 hours is

$$P(X<85)=\Phi(-1)\approx 0.15866,$$

and the probability of lasting more than 130 hours is

$$P(X>130)=1-\Phi(2)\approx 0.02275.$$

These two tails account for the remaining probability:

$$0.15866+0.02275+0.81859=1.00000.$$

Finding a value from a probability

Sometimes the probability is given and the required value is unknown. Suppose you want $x$ such that

$$P(X\le x)=p.$$

First find the standard normal quantile $z_p$ satisfying

$$\Phi(z_p)=p.$$

Then transform back to the original scale:

$$x=\mu+\sigma z_p.$$

For example, the 95th percentile of $X\sim\mathcal{N}(100,15^2)$ uses $z_{0.95}\approx1.64485$:

$$x=100+15(1.64485) \approx124.67.$$

Approximately 95% of components last no longer than 124.67 hours.

For a central 95% interval, use $z_{0.975}\approx1.96$:

$$100\pm1.96(15) =100\pm29.4,$$

so the interval is approximately

$$[70.6,129.4].$$

This is a probability interval for an individual normally distributed observation. It is not automatically a confidence interval for a population mean; that requires a separate sampling model.

Using software or a calculator

A normal-probability calculator usually asks for the mean, standard deviation, and one or two bounds. For the worked example, enter:

• mean: $100$
• standard deviation: $15$
• lower bound: $85$
• upper bound: $130$

The result should be about $0.8186$.

In software, the same calculation can be expressed using a normal CDF. In Python with SciPy, for example:

from scipy.stats import norm

mu = 100
sigma = 15
probability = norm.cdf(130, loc=mu, scale=sigma) - norm.cdf(85, loc=mu, scale=sigma)
print(probability)  # approximately 0.81859

Common mistakes

1. Using the variance as the standard deviation. If the model is written as $\mathcal{N}(\mu,\sigma^2)$, the second parameter is often the variance, while the standardization formula needs $\sigma=\sqrt{\sigma^2}$.

2. Forgetting to subtract the mean. The correct numerator is $x-\mu$, not just $x$.

3. Using the wrong tail. $\Phi(z)$ is the area to the left. For the area to the right, calculate $1-\Phi(z)$.

4. Subtracting raw values instead of CDF values. For an interval, calculate the CDF at the upper endpoint minus the CDF at the lower endpoint.

5. Rounding too early. Keep several digits in the $z$-scores or calculator results, then round the final probability.

6. Assuming every variable is normal. The method is exact only when the normal model is appropriate. For a sample mean, normal calculations may also be justified approximately by the central limit theorem, depending on the sample size and the underlying distribution.