Standard Normal Distribution

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1, written N(0, 1). Its values are z-scores. A z-table or a calculator gives the area to the left of any z. From that, you can find the area to the right of z or between two z-scores.

Before this lesson: Normal Distribution, Empirical Rule and Percentiles and Z-Scores

Why one curve is enough

Every normal curve has the same shape. A z-score tells how many standard deviations a value is from its mean. When you change every value to a z-score, any normal distribution becomes the same curve: N(0, 1).

So one table of areas works for every normal distribution. The standard normal curve is a density curve, so its areas are probabilities. Its center, z = 0, is both the mean and the median, and its standard deviation is 1.

How to read a z-table

A z-table lists the area to the left of z, which is P(Z \lt z). The rows give z to one decimal place. The columns give the second decimal place.

1. Split z into its row and column. For z = 1.25, the row is 1.2 and the column is .05.
2. Read the area where the row and column meet: 0.8944.
3. For a right area, subtract from 1: P(Z \gt z) = 1 - P(Z \lt z).
4. For an area between two z-scores, subtract the two left areas: P(a \lt Z \lt b) = P(Z \lt b) - P(Z \lt a).

Part of the z-table (areas to the left of z)

For a negative z, the row is negative too, and the column still adds to the size of z. For z = −0.68, use row −0.6 and column .08.

Calculator: on a TI-84, press 2nd, VARS (DISTR) and choose normalcdf. Enter the lower bound, the upper bound, μ = 0 and σ = 1. For a left area, use a very small lower bound such as −1E99. In Desmos, type normaldist(0, 1).cdf(a, b) for the area between a and b.

Standard normal distribution examples

Example 1: Area to the left of z

Find P(Z \lt 1.25).

1.25 = 1.2 + 0.05

P(Z \lt 1.25) = 0.8944

Answer: P(Z \lt 1.25) = 0.8944. Calculator check: normalcdf(−1E99, 1.25, 0, 1) ≈ 0.8944.

Example 2: Area to the right of z

Find P(Z \gt -0.68).

P(Z \lt -0.68) = 0.2483

P(Z \gt -0.68) = 1 - 0.2483

= 0.7517

Answer: P(Z \gt -0.68) = 0.7517. This makes sense: −0.68 is left of the center, so more than half of the area is to its right.

Example 3: Area between two z-scores

Find P(-1.30 \lt Z \lt 0.85).

P(Z \lt 0.85) = 0.8023

P(Z \lt -1.30) = 0.0968

P(-1.30 \lt Z \lt 0.85) = 0.8023 - 0.0968

= 0.7055

Answer: about 0.7055. On a TI-84, normalcdf(−1.30, 0.85, 0, 1) ≈ 0.7055. In Desmos, normaldist(0, 1).cdf(−1.3, 0.85) gives the same value.

Example 4: Test scores reported as z-scores (word problem)

A state reading test has scores that are approximately normal, and each student’s report gives a z-score. Leo’s z-score is 2.10. What proportion of students scored higher than Leo? About what percentile is he at?

P(Z \lt 2.10) = 0.9821

P(Z \gt 2.10) = 1 - 0.9821

= 0.0179

Answer: about 0.0179, or 1.79%, scored higher. About 98.21% scored lower, so Leo is at about the 98th percentile.

Example 5: Find z from an area

(a) Find the z-score with 90% of the area to its left. (b) Find the z-score that cuts off the top 5%.

0.8997 \text{ at } z = 1.28

0.9015 \text{ at } z = 1.29

1 - 0.05 = 0.95

0.9495 \text{ at } z = 1.64

0.9505 \text{ at } z = 1.65

z \approx 1.645

Answer: (a) z ≈ 1.28; (b) z ≈ 1.645. On a TI-84, invNorm(0.90, 0, 1) ≈ 1.2816 and invNorm(0.95, 0, 1) ≈ 1.6449. The invNorm command always uses the area to the left.

Common mistakes

Related methods

Any normal distribution

For a normal variable with any mean and standard deviation, first change each value to a z-score with z = \dfrac{x - \mu}{\sigma}. Then use the table. That is the next lesson, Normal Probability Calculations.

Check with the empirical rule

The table agrees with the empirical rule. The area between z = −1 and z = 1 is 0.8413 − 0.1587 = 0.6826, close to 68%.

Practice problems

Use the table above or a calculator.

1. Find P(Z \lt 0.85).
2. Find P(Z \gt 1.25).
3. Find P(Z \lt -1.30).
4. Find P(-0.68 \lt Z \lt 1.25).
5. Find P(Z \gt -1.30).
6. Find P(-2 \lt Z \lt 2). Compare it with the empirical rule.
7. Find P(1.20 \lt Z \lt 2.10).
8. Find the z-score with 20% of the area to its left.
9. Find the z-score that cuts off the top 10%.
10. Fitness test results are reported as z-scores. What proportion of students have z-scores between −1 and 1?

P(Z \lt 0.85) = 0.8023

P(Z \lt 1.25) = 0.8944

P(Z \gt 1.25) = 1 - 0.8944

= 0.1056

P(Z \lt -1.30) = 0.0968

P(Z \lt 1.25) = 0.8944

P(Z \lt -0.68) = 0.2483

0.8944 - 0.2483 = 0.6461

P(Z \lt -1.30) = 0.0968

P(Z \gt -1.30) = 1 - 0.0968

= 0.9032

P(Z \lt 2) = 0.9772

P(Z \lt -2) = 0.0228

0.9772 - 0.0228 = 0.9544

P(Z \lt 2.10) = 0.9821

P(Z \lt 1.20) = 0.8849

0.9821 - 0.8849 = 0.0972

0.2005 \text{ at } z = -0.84

0.1977 \text{ at } z = -0.85

1 - 0.10 = 0.90

z \approx 1.28

P(Z \lt 1) = 0.8413

P(Z \lt -1) = 0.1587

0.8413 - 0.1587 = 0.6826

Standard normal distribution FAQ

What is the standard normal distribution?

It is the normal distribution with mean 0 and standard deviation 1. Its values are z-scores, and its areas come from a z-table or a calculator.

How do you read a z-table?

Find the row for the first decimal of z and the column for the second decimal. The number where they meet is the area to the left of z.

What is the area to the left of z = 0?

It is 0.5. The curve is symmetric about 0, so half of the area is on each side.

How do you find the area between two z-scores?

Find the left area for each z-score. Then subtract the smaller one from the larger one.

Summary

• The standard normal distribution is N(0, 1), and its values are z-scores.
• A z-table gives the area to the left of z.
• Right area: 1 minus the left area. Area between: the larger left area minus the smaller one.
• To find z from an area, look for the area inside the table, or use invNorm.
• On a TI-84, normalcdf finds areas. In Desmos, normaldist(0, 1).cdf finds them.

Previous lesson: Empirical Rule

Next lesson: Normal Probability Calculations

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