How to find a probability with the normal distribution

To find a probability with the normal distribution, convert the variable XX to a standard score ZZ and evaluate the cumulative distribution function. This method applies whenever a continuous random variable is known or assumed to follow a normal distribution N(μ,σ2)N(\mu, \sigma^2).

The setup

Identify the population mean μ\mu, the population standard deviation σ\sigma, and the target boundary values xx. Define the required probability bound, choosing among a left-tail P(X<x)P(X < x), a right-tail P(X>x)P(X > x), or an interval P(x1<X<x2)P(x_1 < X < x_2).

The steps

  1. Calculate the z-score for each boundary value using z=xμσz = \frac{x - \mu}{\sigma}.
  2. 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)P(Z < z).
  3. Adjust the cumulative area for your specific bound: use 1P(Z<z)1 - P(Z < z) for right-tail probabilities, or P(Z<z2)P(Z<z1)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 xx close to the mean μ\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 σ2\sigma^2 instead of the standard deviation σ\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 μ=170\mu = 170 cm and σ=10\sigma = 10 cm. Find the probability that a randomly selected person is between 165 cm and 180 cm tall.

Let XN(170,102)X \sim N(170, 10^2). We need to find P(165<X<180)P(165 < X < 180).

Step 1: Calculate z-scores for both boundaries. z1=16517010=0.5z_1 = \frac{165 - 170}{10} = -0.5 z2=18017010=1.0z_2 = \frac{180 - 170}{10} = 1.0

Step 2: Look up cumulative probabilities in the Z-table. P(Z<0.5)=0.3085P(Z < -0.5) = 0.3085 P(Z<1.0)=0.8413P(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)=0.84130.3085=0.5328P(-0.5 < Z < 1.0) = 0.8413 - 0.3085 = 0.5328

The probability is 0.5328.

FAQ

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

See also