Standard Normal Distribution

Learn the standard normal distribution N(0, 1), how to standardize any normal variable and how to read a z-table for left, right and between areas.

Advanced30 min18 problems
Practice Problems

Definition

The standard normal distribution is the normal distribution with mean μ=0\mu = 0 and standard deviation σ=1\sigma = 1. We write it as N(0,1)N(0, 1) and call its variable ZZ.

Standardizing

Every normal variable XX with mean μ\mu and standard deviation σ\sigma turns into the standard normal when you standardize it:
Z=X−μσZ = \frac{X - \mu}{\sigma}
So one table is enough for every normal distribution: first find zz, then look up the area.

The z-table

A z-table (standard normal table) lists the cumulative area to the left of zz:
Φ(z)=P(Z≤z)\Phi(z) = P(Z \le z)
The row gives zz to one decimal (for example 1.21.2), the column gives the second decimal (for example 0.030.03). Where they meet you read Φ(1.23)=0.8907\Phi(1.23) = 0.8907.

Three rules for every area

  • **Left of zz**: read the table, P(Z<z)=Φ(z)P(Z < z) = \Phi(z)
  • **Right of zz**: the total area is 1, so P(Z>z)=1−Φ(z)P(Z > z) = 1 - \Phi(z)
  • **Between aa and bb**: P(a<Z<b)=Φ(b)−Φ(a)P(a < Z < b) = \Phi(b) - \Phi(a)
The curve is symmetric around 0, so Φ(−z)=1−Φ(z)\Phi(-z) = 1 - \Phi(z). For a continuous variable P(Z<z)P(Z < z) and P(Z≤z)P(Z \le z) are the same number.

Try it now

Which mean and standard deviation does the standard normal distribution N(0,1)N(0, 1) have?

Worked Examples

Find P(Z≤1.23)P(Z \le 1.23) for the standard normal variable ZZ.
1

Split zz into the row and the column

1.23=1.2+0.031.23 = 1.2 + 0.03 → Row 1.21.2, column 0.030.03
2

Read the value where the row and column meet

Row 1.21.2, column 0.030.03: 0.89070.8907 → Φ(1.23)=0.8907\Phi(1.23) = 0.8907
3

Check that it makes sense

z=1.23z = 1.23 is a bit more than 1, and Φ(1)=0.8413\Phi(1) = 0.8413 → A value a little above 0.84130.8413 is reasonable

Common Mistakes

Using the table value as the area to the right: saying P(Z>1.5)=0.9332P(Z > 1.5) = 0.9332

Why it's wrong: The z-table always gives the area to the LEFT of zz. An area of 0.93320.9332 to the right of a positive zz is impossible, because more than half of the curve would lie to its right.

Correct: For a right area subtract from 1: P(Z>1.5)=1−0.9332=0.0668P(Z > 1.5) = 1 - 0.9332 = 0.0668.

Thinking a negative zz gives a negative area, or the same area: Φ(−0.75)=−0.7734\Phi(-0.75) = -0.7734 or 0.77340.7734

Why it's wrong: Areas are never negative. A negative zz lies left of the mean, so the area to its left is less than 0.50.5.

Correct: Use symmetry: Φ(−0.75)=1−Φ(0.75)=1−0.7734=0.2266\Phi(-0.75) = 1 - \Phi(0.75) = 1 - 0.7734 = 0.2266.

Subtracting the z-values instead of the areas: P(−1<Z<2)=Φ(2−(−1))=Φ(3)P(-1 < Z < 2) = \Phi(2 - (-1)) = \Phi(3)

Why it's wrong: The table turns z-values into areas. The area between two points is the difference of the two left areas, not the area of the difference.

Correct: P(−1<Z<2)=Φ(2)−Φ(−1)=0.9772−0.1587=0.8185P(-1 < Z < 2) = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185.

Looking up the raw value in the z-table: P(X<60)P(X < 60) read as row 60

Why it's wrong: The table only works for ZZ with mean 0 and standard deviation 1. A raw value must be standardized first.

Correct: Compute z=60−508=1.25z = \frac{60 - 50}{8} = 1.25 first, then read Φ(1.25)=0.8944\Phi(1.25) = 0.8944.

Interactive Visual

Read the z-table: drag z and watch the row and column

z = 1.23z-3-2-10123

μ = 0, σ = 1 · Drag a handle or type x or z.

P(Z ≤ 1.23) ≈ 0.8907 = 89.07%

About 89.07% of all values fall in the shaded region.

z-table: P(Z ≤ z)

z0.010.020.030.040.05
1.00.84380.84610.84850.85080.8531
1.10.86650.86860.87080.87290.8749
1.20.88690.88880.89070.89250.8944
1.30.90490.90660.90820.90990.9115
1.40.92070.92220.92360.92510.9265

Row 1.2 and column 0.03 give z = 1.23, so P(Z ≤ 1.23) = 0.8907.

Drag a handle: the highlighted row and column follow z.

Standard normal: areas between two z-values and the 68-95-99.7 check

34.1%34.1%13.6%13.6%2.1%2.1%z = -1z = 2z-3-2-10123

μ = 0, σ = 1 · Drag a handle or type x or z.

P(-1 ≤ Z ≤ 2) ≈ 0.8186 = 81.86%

About 81.86% of all values fall in the shaded region.

Standardize: X with mean 50 and sd 8, the z axis below

60 · z = 1.25xz26-334-242-1500581662743X

μ = 50, σ = 8 · Drag a handle or type x or z.

z = (x − μ)/σ = (60 − 50)/8 = 1.25

P(X ≤ 60) = P(Z ≤ 1.25) ≈ 0.8944 = 89.44%

About 89.44% of all values fall in the shaded region.

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Practice Problems

18 problems
Problem 1 of 18
Easy
Which mean and standard deviation does the standard normal distribution N(0,1)N(0, 1) have?

Why It Matters

The standard normal is the one curve that every normal problem is turned into:
  • One table for everything: heights, test scores, reaction times and machine errors have different means and spreads, but after standardizing they all use the same z-table
  • Exact answers instead of estimates: the 68-95-99.7 rule only works at whole standard deviations; the table gives the area for any zz, like z=1.23z = 1.23
  • Statistics and science: confidence intervals, significance tests and quality limits (like ±1.96\pm 1.96 or ±3\pm 3) are all read from the standard normal
  • A built-in check: if a table answer is far from 68%, 95% or 99.7% where it should be close, you know a step went wrong
Once you can read the z-table and use the three area rules, every normal probability problem becomes a short routine.

Real World Applications

Quality Control

Factories standardize every measurement and reject parts that land too far from the target, for example more than 2 standard deviations away.

Example:

A part with z=−2.3z = -2.3 is 2.3 standard deviations below the target length, so it is rejected.

1Try It Yourself

A machine's errors, after standardizing, follow the standard normal distribution. A part is rejected when Z<−2Z < -2 or Z>2Z > 2. Table: Φ(2)=0.9772\Phi(2) = 0.9772.

What share of parts is rejected? Give a decimal.

Step 1: Write the mathematical expression

Two equal tails: 2×(1−0.9772)2 \times (1 - 0.9772)

Standardized Test Scores

Some exams report results directly as z-scores, so a student can see at once how far above or below the average they are.

Example:

A reported score of z=0z = 0 means exactly average; z=1z = 1 means one standard deviation above average.

2Try It Yourself

Scores on an exam are reported as z-scores and follow the standard normal distribution. Table: Φ(0.5)=0.6915\Phi(0.5) = 0.6915.

What share of students has a z-score above 0.50.5?

Step 1: Write the mathematical expression

Right area: 1−Φ(0.5)1 - \Phi(0.5)

Weather Forecast Errors

Forecasters check how far their temperature predictions miss. For a well calibrated forecast the error in °C can follow the standard normal distribution: mean 0, standard deviation 1.

Example:

An error of −1.2-1.2 °C means the real temperature was 1.2 degrees lower than predicted.

3Try It Yourself

Next-day forecast errors (in °C) follow the standard normal distribution. Table: Φ(1.5)=0.9332\Phi(1.5) = 0.9332.

What is the probability that a forecast misses by less than 1.5 °C, so −1.5<Z<1.5-1.5 < Z < 1.5?

Step 1: Write the mathematical expression

Between: Φ(1.5)−Φ(−1.5)\Phi(1.5) - \Phi(-1.5)

Key Takeaways

  • 1The standard normal distribution N(0,1)N(0, 1) has mean 0 and standard deviation 1; its variable is called ZZ
  • 2Standardize any normal value with z=x−μσz = \frac{x - \mu}{\sigma}, then use the one z-table
  • 3The z-table gives the area to the LEFT: Φ(z)=P(Z≤z)\Phi(z) = P(Z \le z); row = first decimal, column = second decimal
  • 4Right area: P(Z>z)=1−Φ(z)P(Z > z) = 1 - \Phi(z); between: P(a<Z<b)=Φ(b)−Φ(a)P(a < Z < b) = \Phi(b) - \Phi(a)
  • 5Symmetry: Φ(−z)=1−Φ(z)\Phi(-z) = 1 - \Phi(z) and Φ(0)=0.5\Phi(0) = 0.5
  • 6Check every answer with the 68-95-99.7 rule: for example Φ(1)−Φ(−1)=0.6826≈68%\Phi(1) - \Phi(-1) = 0.6826 \approx 68\%

Frequently Asked Questions

Standardizing shifts the mean to 0 and rescales the spread to 1. The shape stays the same, so the area left of xx for a normal distribution with mean μ\mu and standard deviation σ\sigma equals the area left of z=x−μσz = \frac{x - \mu}{\sigma} in N(0,1)N(0, 1).
Standardizing shifts the mean to 0 and rescales the spread to 1. The shape stays the same, so the area left of xx for a normal distribution with mean μ\mu and standard deviation σ\sigma equals the area left of z=x−μσz = \frac{x - \mu}{\sigma} in N(0,1)N(0, 1).
Use symmetry: Φ(−z)=1−Φ(z)\Phi(-z) = 1 - \Phi(z). For example Φ(−1.2)=1−0.8849=0.1151\Phi(-1.2) = 1 - 0.8849 = 0.1151. Many books print a second table for negative zz; both give the same numbers.
No. ZZ is continuous, so the probability of hitting exactly one value is 0. Both are read from the same table entry.
Most tables stop at about 3.493.49, where Φ\Phi is already 0.99980.9998. Beyond that the left area is practically 1 and the right tail practically 0.

Glossary

Standard normal distribution
The normal distribution with mean 0 and standard deviation 1, written N(0,1)N(0, 1)
Standardize
Turn a value xx into z=x−μσz = \frac{x - \mu}{\sigma}, the number of standard deviations from the mean
z-table
A table of the cumulative areas Φ(z)=P(Z≤z)\Phi(z) = P(Z \le z) to the left of each zz
Cumulative area Φ(z)\Phi(z)
The area under the standard normal curve to the left of zz; it equals the probability P(Z≤z)P(Z \le z)
Tail
The area at one end of the curve, for example to the right of z=2z = 2

Formula Card

Standardizing

z=x−μσz = \frac{x - \mu}{\sigma}

Turns any normal value into a value of $Z$

Left area

P(Z<z)=Φ(z)P(Z < z) = \Phi(z)

Read directly from the z-table

Right area

P(Z>z)=1−Φ(z)P(Z > z) = 1 - \Phi(z)

The total area under the curve is 1

Between

P(a<Z<b)=Φ(b)−Φ(a)P(a < Z < b) = \Phi(b) - \Phi(a)

Larger left area minus smaller left area

Symmetry

Φ(−z)=1−Φ(z)\Phi(-z) = 1 - \Phi(z)

For negative z-values

Today's challengeOne problem, three tries

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