Normal distributions

This is about probability of measuring some continuous quantity, such as time, length, and mass, to be within some interval.

The properties on this page only applies to normal distribution

Updated 2026-05-13 on GDC (calculator) usage

When to Use
Definition
Mean = Median = Mode
Empirical rule and $z$ -score
Calculator
Examples
Tips

When to Use

If it exhibits the following symptoms, normal distribution may be for you

question mentions normal distribution, and
measurement is continuous (time, length, mass etc), and one of
a measurement of sample size 1 (eg a random measurement from the population), or
finding a fraction or percentage of the population

Just because question mentions normal distribution does not mean you will only use normal. It is often the case that a question starts out normal, only to shift into binomial distribution regarding $k$ successes out of $n$ tries.

Definition

A normally distributed random variable $X$ is notated as

X \sim \text{N}(\mu, \sigma^2)

with expected value (or mean) $\mu$ and standard deviation $\sigma > 0$ or variance $\sigma^2$ .

English: A selection of Normal Distribution Probability Density Functions (PDFs). Both the mean, μ, and variance, σ², are varied. The key is given on the graph. — Wikipedia: by Inductiveload. Smaller σ is narrower and taller. Larger σ is wider and shorter.

The area under the curve from $x = a$ to $x = b$ is $\text P(a \leq X \leq b)$ . The probability of any particular value is $0$ .

\text{P}(X = a) = 0

\text{P}(a \leq X \leq b) = \text{P}(a < X < b)

For values within distance $d$ of $k$ , you may see

\text{P}(\lvert X - k \rvert < d) = \text{P}(k - d < X < k + d)

For values outside of distance $d$ of $k$ , you may see

\begin{align*} &\,\text{P}(\lvert X - k \rvert > d) \\ =&\, \text{P}(X < k - d) + \text{P}(X > k + d) \end{align*}

Mean = Median = Mode

For a normal distribution, $\mu$ is the mean, the median, and the mode. This means

\text{P}(X < \mu) = \text{P}(X > \mu) = 0.5

This also means that normal distribution is symmetric about the mean.

Other properties based on the symmetry include, for some $0 < j < k$

\begin{align*} \text{P}(\mu - k < X < \mu) &= \text{P}(\mu < X < \mu + k) \\ &= 0.5 - \text{P}(X < \mu - k) \end{align*}

\begin{align*} &\,\text{P}(\mu - k < X < \mu - j) \\ =&\, \text{P}(\mu + j < X < \mu + k) \end{align*}

Empirical rule and $z$ -score

The empirical rule, or $68-95-99.7$ rule, states that, for a normal distribution, regardless of the values of $\mu$ or $\sigma$

\text{P}(\mu - \sigma < X < \mu + \sigma) = \text{P}(\lvert X - \mu \rvert < \sigma) \approx 0.68

\text{P}(\mu - 2\sigma < X < \mu + 2\sigma) = \text{P}(\lvert X - \mu \rvert < 2\sigma) \approx 0.95

\text{P}(\mu - 3\sigma < X < \mu + 3\sigma) = \text{P}(\lvert X - \mu \rvert < 3\sigma) \approx 0.997

For a single measurement, there is a $68\%$ chance it is within $1$ standard deviation of the mean, $95\%$ chance of within $2\sigma$ , and $99.7\%$ chance of within $3\sigma$ .

The above values must be memorized for Paper 1.

This prompts the definition of the $z$ -score

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

The $z$ -score is the number of standard deviations that a value $x$ is more than the mean. Negative $z$ -scores mean the value is that many standard deviations below the mean.

Then all calculations can simply refer to the standardized normal distribution $\text{N}(0, 1)$ .

$z$ -scores allow for solving for $\mu$ or $\sigma$ when given the probability. In summary, they extend the power of invNorm or InvN of your graphing calculator.

A sketch of the derivation is provided in the discussion on definite integrals.

Calculator

This is about graphing display calculator (GDC) use. Arguments in $\color{grey}\text{grey}$ are optional. By default, $\mu =0$ , $\sigma = 1$ .

	menu	P(a ≤ X ≤ b)	x such that P(X ≤ x) = cdf
TI-84	2nd vars	2.normalcdf(a, b $,\color{grey}\mu, \sigma$ )	3.invNorm(cdf $,\color{grey}\mu, \sigma, \text{tail}$ )
Nspire	menu 6.statistics, then 5.distributions	2.normalcdf(a, b $,\color{grey}\mu, \sigma$ )	3.invNorm(cdf $,\color{grey}\mu, \sigma$ )
Casio*	apps `I.DIST`, or `2.STAT`, or `1.Run/Mat` OPTN `stat`	Ncd(a, b $\color{grey}, \sigma, \mu$ )	InvN( $\color{grey}\text{ -1 or 0 or 1, }$ p $\color{grey},\sigma, \mu$ )

Casio: Use 2.Statistics or I.Distribution apps by default. Use 2. Stat if needing InvN. Only use the 1.Run/Mat to solve for unknown $\mu$ or $\sigma$ by graphing or nSolve, as an alternative to $z$ -scores.

By default, inverse normal uses lower tail, to find P(X ≤ x) = cdf. TI-84 Plus CE and Casio can find lower (LEFT, -1), central (CENTER, 0) or upper (RIGHT, 1) tails, which can only be specified if you include both $\sigma$ and $\mu$ . Eg upper would be x such that P(X ≥ x) = cdf. On Casio for central, only the lower limit is returned; the upper limit would be $2\mu - \text{ans}$

For probability where the interval involves $-\infty$ or $\infty$ , use $-1000$ or $1000$ , or any arbitrary values more than $5$ standard deviations away from the mean in the appropriate direction.

Examples

On TI-84 Plus and TI-Nspire, the inverse normal function invNorm returns $x$ , such that $\text{P}(X < x) = p$ for some known distribution and probability $p$ . On Casio and other brands, there is often an option to solve $\text{P}(X > x) = p$ as well. We shall assume we can only solve the first case on the calculator.

Example: Let $X \sim \text{N}(50, 16)$ , find $x$ such that $\text{P}(40 < X < x) = 0.200$ .

The normal distribution has mean $50$ and standard deviation $4$ . We have

\begin{align*} &\,\text{P}(X < 40) + \text{P}(40 < X < x) \\ =&\, \text{P}(X < x) \end{align*}

Then,

\begin{align*} \text{P}(40 < X < x) &= 0.200 \\ \text{P}(X < x) - \text{P}(X < 0.40) &= 0.200 \\ \text{P}(X < x) - 0.00621\dots &= 0.200 \\ \text{P}(X < x) &= 0.20621\dots \\ x &\approx 46.7 \qed \end{align*}

Example: The percentage by volume of potato chips in a bag is normally distributed. The company wants $97\%$ of bags to contain at least $65\%$ chips, and $1\%$ of bags to contain at least $70\%$ chips. To three significant figures, find the mean and standard deviation to pack the least amount of chips.

Let $Y$ be the percentage of chips in the bag.

At the minimum amount of chips packed,

\text{P}(Y > 65) = 0.97 \text{ and } {\text{P}(Y > 70) = 0.01}

The endpoints of the standardized normal distribution with left-tail probabilities $0.03$ and $0.99$ are $-1.88079$ and $2.32635$ , respectively.

Using $z$ -scores,

\begin{align*} \frac{65 - \mu}{\sigma} &= -1.88079\dots \\ \frac{70 - \mu}{\sigma} &= 2.32635\dots \end{align*}

Most calculators require the equations in the form

\begin{align*} -1.88079\sigma + \mu &= 65 \\ 2.32635\sigma + \mu &= 70 \end{align*}

before finding $\mu = 67.2, \sigma = 1.19 \qed$

The bags should be filled with mean $67.2\%$ and standard deviation $1.19\%$ .

Tips

Draw a bell curve. Label the mean and the desired interval(s).
Which way is the inequality sign?
Does it want probability of being inside or outside some interval?
Does it want a minimum or a maximum?
Math notation uses variance but calculators typically use standard deviation.
Write down the calculator command you used, so you can easily check your work later.
The intersection of two events is their area(s) of overlap. Conditional probability is a ratio (fraction) of areas. See also probability formulas.

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