Standard Normal Distribution

The **standard normal distribution** is the normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$. We write it as $N(0, 1)$ and call its variable $Z$. ## Standardizing Every normal variable $X$ with mean $\mu$ and standard deviation $\sigma$ turns into the standard normal when you **standardize** it: $$Z = \frac{X - \mu}{\sigma}$$ So one table is enough for every normal distribution: first find $z$, then look up the area. ## The z-table A **z-table** (standard normal table) lists the **cumulative area to the left** of $z$: $$\Phi(z) = P(Z \le z)$$ The **row** gives $z$ to one decimal (for example $1.2$), the **column** gives the second decimal (for example $0.03$). Where they meet you read $\Phi(1.23) = 0.8907$. ## Three rules for every area - **Left of $z$**: read the table, $P(Z < z) = \Phi(z)$ - **Right of $z$**: the total area is 1, so $P(Z > z) = 1 - \Phi(z)$ - **Between $a$ and $b$**: $P(a < Z < b) = \Phi(b) - \Phi(a)$ The curve is symmetric around 0, so $\Phi(-z) = 1 - \Phi(z)$. For a continuous variable $P(Z < z)$ and $P(Z \le z)$ are the same number.

Reading the z-table

Find $P(Z \le 1.23)$ for the standard normal variable $Z$.

Split $z$ into the row and the column: $1.23 = 1.2 + 0.03$ = Row $1.2$, column $0.03$

Read the value where the row and column meet: Row $1.2$, column $0.03$: $0.8907$ = $\Phi(1.23) = 0.8907$

Check that it makes sense: $z = 1.23$ is a bit more than 1, and $\Phi(1) = 0.8413$ = A value a little above $0.8413$ is reasonable

Answer: $P(Z \le 1.23) = 0.8907$, so about 89.07% of the area lies to the left of $z = 1.23$.

Area to the right

Find $P(Z > 1.5)$.

Read the left area from the table: Row $1.5$, column $0.00$ = $\Phi(1.5) = 0.9332$

Subtract from the total area 1: $P(Z > 1.5) = 1 - 0.9332$ = $0.0668$

Check with the 68-95-99.7 rule: Above $z = 1$ lies 16%, above $z = 2$ lies 2.5% = $0.0668$ (6.68%) lies between them

Answer: $P(Z > 1.5) = 1 - 0.9332 = 0.0668$

A negative z and symmetry

Find $P(Z < -0.75)$ using only the positive part of the table: $\Phi(0.75) = 0.7734$.

Use symmetry: the left tail at $-0.75$ equals the right tail at $0.75$: $P(Z < -0.75) = P(Z > 0.75)$ = $\Phi(-0.75) = 1 - \Phi(0.75)$

Subtract the table value from 1: $1 - 0.7734$ = $0.2266$

Check the size: $-0.75$ is left of the mean, so the area must be less than $0.5$ = $0.2266 < 0.5$, it fits

Answer: $P(Z < -0.75) = 1 - 0.7734 = 0.2266$

Area between two z-values

Find $P(-1 < Z < 2)$. Table: $\Phi(1) = 0.8413$, $\Phi(2) = 0.9772$.

Left area up to the upper bound: $\Phi(2)$ = $0.9772$

Left area up to the lower bound (symmetry): $\Phi(-1) = 1 - 0.8413$ = $0.1587$

Subtract the two areas: $0.9772 - 0.1587$ = $0.8185$

Check with the 68-95-99.7 rule: From $-1$ to $0$: 34%, from $0$ to $2$: 47.5% = $34\% + 47.5\% = 81.5\%$, close to $0.8185$

Answer: $P(-1 < Z < 2) = 0.9772 - 0.1587 = 0.8185$

Standardizing any normal variable

$X$ is normal with $\mu = 50$ and $\sigma = 8$. Find $P(X < 60)$. Table: $\Phi(1.25) = 0.8944$.

Standardize the bound: $z = \frac{60 - 50}{8} = \frac{10}{8}$ = $z = 1.25$

Rewrite the question with $Z$: $P(X < 60) = P(Z < 1.25)$ = A left area

Read the table: Row $1.2$, column $0.05$ = $0.8944$

Check with the 68-95-99.7 rule: $\Phi(1) \approx 0.84$ and $\Phi(2) \approx 0.975$ = $0.8944$ lies between them

Answer: $P(X < 60) = P(Z < 1.25) = 0.8944$

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 $z$, like $z = 1.23$ - **Statistics and science**: confidence intervals, significance tests and quality limits (like $\pm 1.96$ or $\pm 3$) 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.

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

Why: The z-table always gives the area to the LEFT of $z$. An area of $0.9332$ to the right of a positive $z$ is impossible, because more than half of the curve would lie to its right.

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

Mistake: Thinking a negative $z$ gives a negative area, or the same area: $\Phi(-0.75) = -0.7734$ or $0.7734$

Why: Areas are never negative. A negative $z$ lies left of the mean, so the area to its left is less than $0.5$.

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

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

Why: 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: $P(-1 < Z < 2) = \Phi(2) - \Phi(-1) = 0.9772 - 0.1587 = 0.8185$.

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

Why: The table only works for $Z$ with mean 0 and standard deviation 1. A raw value must be standardized first.

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

Quality Control

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

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

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.

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

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.

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

The standard normal distribution $N(0, 1)$ has mean 0 and standard deviation 1; its variable is called $Z$

Standardize any normal value with $z = \frac{x - \mu}{\sigma}$, then use the one z-table

The z-table gives the area to the LEFT: $\Phi(z) = P(Z \le z)$; row = first decimal, column = second decimal

Right area: $P(Z > z) = 1 - \Phi(z)$; between: $P(a < Z < b) = \Phi(b) - \Phi(a)$

Symmetry: $\Phi(-z) = 1 - \Phi(z)$ and $\Phi(0) = 0.5$

Check every answer with the 68-95-99.7 rule: for example $\Phi(1) - \Phi(-1) = 0.6826 \approx 68\%$

Q: Why does one table work for every normal distribution?

A: Standardizing shifts the mean to 0 and rescales the spread to 1. The shape stays the same, so the area left of $x$ for a normal distribution with mean $\mu$ and standard deviation $\sigma$ equals the area left of $z = \frac{x - \mu}{\sigma}$ in $N(0, 1)$.

Q: My table has no negative z-values. What do I do?

A: Use symmetry: $\Phi(-z) = 1 - \Phi(z)$. For example $\Phi(-1.2) = 1 - 0.8849 = 0.1151$. Many books print a second table for negative $z$; both give the same numbers.

Q: Is there a difference between $P(Z < z)$ and $P(Z \le z)$?

A: No. $Z$ is continuous, so the probability of hitting exactly one value is 0. Both are read from the same table entry.

Q: What if my $z$ is larger than the table, like $z = 4$?

A: Most tables stop at about $3.49$, where $\Phi$ is already $0.9998$. Beyond that the left area is practically 1 and the right tail practically 0.

Exit Presentation

Standard Normal Distribution

1 / 14

Probability / Probability Distributions

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 min

Use arrow keys or spacebar to navigate