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Standard Normal Distribution — Definition, Formula & Examples

The standard normal distribution is a normal (bell-shaped) distribution with a mean of 0 and a standard deviation of 1. Any normal distribution can be converted to it using z-scores, making it the universal reference for finding probabilities.

A continuous probability distribution Z∼N(0,1)Z \sim N(0,1) whose probability density function is f(z)=12π e−z2/2f(z) = \frac{1}{\sqrt{2\pi}}\,e^{-z^2/2} for z∈(−∞,∞)z \in (-\infty, \infty). Every normally distributed random variable X∼N(μ,σ2)X \sim N(\mu, \sigma^2) can be standardized to this distribution via the transformation Z=X−μσZ = \frac{X - \mu}{\sigma}.

Key Formula

z=x−μσz = \frac{x - \mu}{\sigma}
Where:
  • zz = The z-score (number of standard deviations from the mean)
  • xx = The raw data value you want to standardize
  • μ\mu = The mean of the original normal distribution
  • σ\sigma = The standard deviation of the original normal distribution

How It Works

To use the standard normal distribution, you first convert a raw data value xx into a z-score using the formula z=x−μσz = \frac{x - \mu}{\sigma}. The z-score tells you how many standard deviations xx is above or below the mean. You then look up the z-score in a z-table (or use a calculator) to find the cumulative probability — the area under the curve to the left of that z-value. This area equals the probability that a randomly chosen value falls below xx. Because the total area under the curve is 1, you can also find right-tail or between-two-values probabilities by subtracting from 1 or combining areas.

Worked Example

Problem: Test scores are normally distributed with a mean of 500 and a standard deviation of 100. What is the probability that a randomly selected student scores below 680?
Step 1: Identify the given values: mean, standard deviation, and the score of interest.
μ=500,σ=100,x=680\mu = 500, \quad \sigma = 100, \quad x = 680
Step 2: Convert the raw score to a z-score.
z=680−500100=180100=1.80z = \frac{680 - 500}{100} = \frac{180}{100} = 1.80
Step 3: Look up z = 1.80 in a standard normal z-table. The table gives the cumulative area to the left of z = 1.80.
P(Z<1.80)=0.9641P(Z < 1.80) = 0.9641
Answer: There is approximately a 96.41% probability that a randomly selected student scores below 680.

Another Example

Problem: Using the same test (mean 500, standard deviation 100), what is the probability a student scores between 400 and 600?
Step 1: Find the z-score for 400.
z1=400−500100=−1.00z_1 = \frac{400 - 500}{100} = -1.00
Step 2: Find the z-score for 600.
z2=600−500100=1.00z_2 = \frac{600 - 500}{100} = 1.00
Step 3: Use the z-table to find cumulative probabilities, then subtract to get the area between.
P(−1.00<Z<1.00)=P(Z<1.00)−P(Z<−1.00)=0.8413−0.1587=0.6826P(-1.00 < Z < 1.00) = P(Z < 1.00) - P(Z < -1.00) = 0.8413 - 0.1587 = 0.6826
Answer: About 68.26% of students score between 400 and 600. This illustrates the well-known 68-95-99.7 rule: roughly 68% of data falls within one standard deviation of the mean.

Visualization

Why It Matters

The standard normal distribution is central to AP Statistics and introductory college statistics courses. It underpins hypothesis testing, confidence intervals, and quality control in manufacturing — whenever you ask whether an observed result is unusually far from what's expected. SAT and ACT scores, medical measurements, and financial models all rely on z-scores derived from this distribution.

Common Mistakes

Mistake: Forgetting that the z-table gives the area to the LEFT of a z-score, then reporting that value as a right-tail probability.
Correction: For the probability above a z-value, subtract the table value from 1: P(Z>z)=1−P(Z<z)P(Z > z) = 1 - P(Z < z).
Mistake: Mixing up the sign of the z-score or subtracting in the wrong order (μ−x\mu - x instead of x−μx - \mu).
Correction: Always compute z=x−μσz = \frac{x - \mu}{\sigma}. A value below the mean should give a negative z-score, and a value above the mean should give a positive z-score.

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