Z-Scores and the Standard Normal Distribution

The 68–95–99.7 rule only works for values exactly $1$, $2$, or $3$ standard deviations from the mean. What about a height that’s $1.75$ standard deviations above? A z-score turns any value into “how many standard deviations from the mean,” so every normal distribution can be compared on the same scale and every probability can be found from one table. It’s also how you compare apples and oranges, like a mark in math with a mark in chemistry.

Key ideas

The z-score

The z-score of a value $x$ tells you how many standard deviations it is from the mean:

$$z = \frac{x - \mu}{\sigma}$$

• $z \gt 0$: above the mean. $z \lt 0$: below the mean. $z = 0$: exactly at the mean.
• $z = 1.5$ means ”$1.5$ standard deviations above the mean.”
• For a sample, use $z = \dfrac{x - \bar{x}}{s}$.

To go back from a z-score to a value, rearrange:

$$x = \mu + z\sigma$$

The standard normal distribution

If $X \sim N(\mu, \sigma^2)$, then the z-scores $Z = \dfrac{X - \mu}{\sigma}$ follow the standard normal distribution:

$$Z \sim N(0, 1)$$

It has mean $0$ and standard deviation $1$. Every normal probability question can be turned into a question about $Z$.

Finding probabilities

The area to the left of $z$, $P(Z \lt z)$, is what a z-table (standard normal table) lists. From it you can get any area:

You want	Use
$P(Z \lt a)$	the table value for $a$
$P(Z \gt a)$	$1 - P(Z \lt a)$
$P(a \lt Z \lt b)$	$P(Z \lt b) - P(Z \lt a)$

Since $X$ is continuous, $\lt$ and $\le$ give the same answer.

Technology can skip the z-score step:

Tool	Area between $a$ and $b$	Area to the left of $x$	Inverse: value with area $p$ to its left
TI-83/84	normalcdf(a, b, μ, σ)	normalcdf(-1E99, x, μ, σ)	invNorm(p, μ, σ)
Spreadsheet	=NORM.DIST(b,μ,σ,TRUE)-NORM.DIST(a,μ,σ,TRUE)	=NORM.DIST(x,μ,σ,TRUE)	=NORM.INV(p,μ,σ)

For the standard normal, =NORM.S.DIST(z,TRUE) and =NORM.S.INV(p) work too.

The answers on this page are given to $4$ decimal places, from technology. A z-table rounds $z$ to $2$ decimals, so your table answer may differ slightly in the last decimal place. That’s fine.

Percentiles

The $k$th percentile is the value with $k\%$ of the data below it. If a value’s area to the left is $0.9032$, it’s at about the $90$th percentile. (More on percentiles for data sets in quartiles and percentiles.)

Working backwards

To find the value $x$ that has a given area $p$ to its left:

1. Find $z$ with $P(Z \lt z) = p$: look inside the table for the area closest to $p$, or use invNorm(p).
2. Convert back: $x = \mu + z\sigma$.

Worked examples

Example 1: Comparing marks

Maya scored $82$ on a math test (class mean $70$, standard deviation $8$) and $88$ on a chemistry test (class mean $80$, standard deviation $6$). On which test did she do better compared with her class?

Solution.

$$z_{\text{math}} = \frac{82 - 70}{8} = 1.5, \qquad z_{\text{chem}} = \frac{88 - 80}{6} \approx 1.33$$

Her math mark is $1.5$ standard deviations above the class mean, compared with about $1.33$ for chemistry. So she did better in math relative to her class, even though her chemistry mark is higher.

Example 2: Area to the right

The heights of Grade 12 students are normally distributed, $X \sim N(172, 8^2)$, in centimetres. What proportion of students are taller than $186$ cm?

Solution. Standardize:

$$z = \frac{186 - 172}{8} = 1.75$$

The table (or technology) gives $P(Z \lt 1.75) = 0.9599$. You want the area to the right:

$$P(X \gt 186) = P(Z \gt 1.75) = 1 - 0.9599 = 0.0401$$

For $X \sim N(172, 8^2)$, the shaded area is $P(X \gt 186) = P(Z \gt 1.75) \approx 0.0401$.

About $4\%$ of students are taller than $186$ cm. On a TI-84: normalcdf(186, 1E99, 172, 8) $\approx 0.0401$.

Since $0.9599$ of students are shorter, a height of $186$ cm is at about the $96$th percentile.

Example 3: Area between two values

For the same heights, find $P(160 \lt X \lt 182)$.

Solution.

$$z_1 = \frac{160 - 172}{8} = -1.5, \qquad z_2 = \frac{182 - 172}{8} = 1.25$$ $$\begin{aligned} P(160 \lt X \lt 182) &= P(Z \lt 1.25) - P(Z \lt -1.5) \\ &= 0.8944 - 0.0668 \\ &= 0.8276 \end{aligned}$$

Technology gives $0.8275$, because it subtracts the unrounded areas ($0.89435\ldots - 0.06681\ldots$). Either answer is fine. About $83\%$ of students are between $160$ cm and $182$ cm.

Example 4: Working backwards to a percentile

How tall must a student be to be at the $90$th percentile?

Solution. Find $z$ with $P(Z \lt z) = 0.90$. In the table, the closest area is $0.8997$, at $z = 1.28$. (Technology: invNorm(0.90) $\approx 1.2816$.)

$$x = \mu + z\sigma = 172 + 1.2816(8) \approx 182.25 \text{ cm}$$

Or directly: invNorm(0.90, 172, 8) $\approx 182.25$. (With the table’s $z = 1.28$, you’d get $182.24$ cm, which is just as good.)

A student about $182.3$ cm tall is taller than $90\%$ of Grade 12 students.

Common mistakes

Using the table value when you want the area to the right. The table gives the area to the left. For “more than” or “greater than”, subtract from $1$. A quick sketch with the region shaded catches this every time.

Dividing by the variance. In $N(172, 64)$, divide by $\sigma = 8$, not $64$.

Getting the sign of z wrong. A value below the mean has a negative z-score, and its left-area is less than $0.5$. If $x \lt \mu$ but you get an area bigger than $0.5$ to the left, check your sign.

Looking up a probability as if it were a z-score. When working backwards, you know the area and need $z$, so search inside the body of the table, not down the side.

Forgetting to convert back. After finding $z$ for a percentile, you still need $x = \mu + z\sigma$ to answer in the original units.

Practice

1. (Warm-up) For a distribution with $\mu = 65$ and $\sigma = 5$, find the z-score of each value: $75$, $58$, $65$, $72.5$.

Solution

$$\frac{75 - 65}{5} = 2, \quad \frac{58 - 65}{5} = -1.4, \quad \frac{65 - 65}{5} = 0, \quad \frac{72.5 - 65}{5} = 1.5$$

2. (Warm-up) For $Z \sim N(0, 1)$, find:

• (a) $P(Z \lt 1.2)$
• (b) $P(Z \gt -0.5)$

Solution

(a) $P(Z \lt 1.2) = 0.8849$

(b) $P(Z \gt -0.5) = 1 - P(Z \lt -0.5) = 1 - 0.3085 = 0.6915$

3. (Warm-up) For the heights $X \sim N(172, 8^2)$, what height has a z-score of $-2.5$?

Solution

$$x = 172 + (-2.5)(8) = 152 \text{ cm}$$

4. (Core) The masses of bags of flour are normally distributed with $\mu = 1010$ g and $\sigma = 8$ g. The label says $1000$ g. What proportion of bags are under the labelled mass?

Solution

$$z = \frac{1000 - 1010}{8} = -1.25$$$$P(X \lt 1000) = P(Z \lt -1.25) = 0.1056$$

About $10.6\%$ of bags are under $1000$ g.

5. (Core) A student’s walk to school takes $X \sim N(30, 5^2)$ minutes. Find the probability that a walk takes between $24$ and $38$ minutes.

Solution

$$z_1 = \frac{24 - 30}{5} = -1.2, \qquad z_2 = \frac{38 - 30}{5} = 1.6$$$$P(24 \lt X \lt 38) = P(Z \lt 1.6) - P(Z \lt -1.2) = 0.9452 - 0.1151 = 0.8301$$

6. (Core) Scores on a provincial math assessment are normally distributed with $\mu = 68$ and $\sigma = 10$. Jordan scores $81$. At what percentile is Jordan’s score?

Solution

$$z = \frac{81 - 68}{10} = 1.3, \qquad P(Z \lt 1.3) = 0.9032$$

Jordan scored higher than about $90\%$ of students: about the $90$th percentile.

7. (Core) On the same assessment, the top $5\%$ of students receive an award. What is the lowest score that earns an award?

Solution

The top $5\%$ means $95\%$ are below the cut-off. Find $z$ with $P(Z \lt z) = 0.95$: $z \approx 1.6449$ (the table gives about $1.645$).

$$x = 68 + 1.6449(10) \approx 84.45$$

A score of about $84.45$ or higher earns an award (so $85$ or higher, if scores are whole numbers). (Technology: invNorm(0.95, 68, 10) $\approx 84.45$.)

8. (Challenge) The masses of eggs from a farm are normally distributed with a mean of $58$ g. $10\%$ of eggs are heavier than $64$ g. Find the standard deviation.

Solution

$10\%$ above $64$ g means $90\%$ below, so $64$ g has $z$ with $P(Z \lt z) = 0.90$: $z \approx 1.2816$.

$$1.2816 = \frac{64 - 58}{\sigma} \quad\Rightarrow\quad \sigma = \frac{6}{1.2816} \approx 4.68 \text{ g}$$

9. (Challenge) Scores on a test are normally distributed. $20\%$ of students score below $50$, and $10\%$ score above $80$. Find the mean and standard deviation.

Solution

$P(Z \lt z) = 0.20$ gives $z \approx -0.8416$. $P(Z \lt z) = 0.90$ gives $z \approx 1.2816$. Using $x = \mu + z\sigma$:

$$\begin{aligned} \mu - 0.8416\sigma &= 50 \\ \mu + 1.2816\sigma &= 80 \end{aligned}$$

Subtract: $2.1232\sigma = 30$, so $\sigma \approx 14.13$. Then $\mu = 50 + 0.8416(14.13) \approx 61.89$.

Check: $P(X \lt 50)$ with $N(61.89, 14.13^2)$ is $0.2000$ and $P(X \gt 80)$ is $0.1000$. ✓