The normal distribution

The normal distribution is a continuous probability model with a symmetric, bell-shaped density curve. It is used for variables whose values cluster around a central value, with increasingly extreme values becoming increasingly rare.

If a continuous random variable $X$ has mean $\mu$ and variance $\sigma^2$, write

$$\boxed{X\sim\operatorname{N}(\mu,\sigma^2)}.$$

The second parameter is the variance, not the standard deviation. Thus

$$X\sim\operatorname{N}(50,16)$$

means that $\mu=50$ and $\sigma=4$.

This lesson explains the model, probability calculations, inverse problems, parameter estimation and the normal approximation to a binomial distribution.

Prerequisites

You should be able to:

• use complements and combine intervals from probability;
• interpret the mean, variance and standard deviation from averages and spread;
• distinguish continuous and discrete random variables;
• use the binomial distribution;
• solve simultaneous equations and use inverse calculator functions.

Shape and parameters

Every normal density curve has these properties:

• it is symmetric about $x=\mu$;
• its mean, median and mode are all $\mu$;
• the total area under the curve is $1$;
• the curve approaches, but never meets, the horizontal axis;
• its points of inflection are at $\mu-\sigma$ and $\mu+\sigma$.

The mean $\mu$ controls the location of the curve. Increasing $\mu$ translates the curve to the right without changing its shape.

The standard deviation $\sigma$ controls the spread. A larger $\sigma$ produces a wider, lower curve because the total area must remain $1$.

The density function is

$$f(x)=\frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right),$$

but A-level probability calculations use calculator distribution functions rather than direct integration of this expression.

Area represents probability

For a continuous random variable, probability is an area under the density curve, not the height of the curve. A density value can even exceed $1$, provided the total area is $1$.

The empirical rule

For every normal distribution, approximately

$$\begin{aligned} \operatorname{P}(\mu-\sigma<X<\mu+\sigma)&\approx0.6827,\\ \operatorname{P}(\mu-2\sigma<X<\mu+2\sigma)&\approx0.9545,\\ \operatorname{P}(\mu-3\sigma<X<\mu+3\sigma)&\approx0.9973. \end{aligned}$$

These values are useful for judging whether a calculator answer is plausible. They should not replace a calculator when an accurate probability is required.

Worked example 1: reading the parameters

The masses $M$ grams of packets are modelled by

$$M\sim\operatorname{N}(250,6.25).$$

State the mean and standard deviation, and estimate the central interval containing about $95\%$ of packet masses.

The mean is

$$\mu=250.$$

Since $\sigma^2=6.25$,

$$\sigma=\sqrt{6.25}=2.5.$$

About $95\%$ of values lie within two standard deviations of the mean:

$$250-2(2.5)<M<250+2(2.5).$$

Therefore the estimated interval is

$$\boxed{245<M<255}.$$

Self-check 1

Suppose $T\sim\operatorname{N}(18,9)$. State the axis of symmetry and the coordinates of the two points of inflection.

Answer

Here $\mu=18$ and $\sigma=\sqrt9=3$. The axis of symmetry is

$$\boxed{x=18},$$

and the points of inflection occur when $x=18\pm3$, so their $x$ coordinates are

$$\boxed{15\text{ and }21}.$$

Continuous probabilities

Since a single point has zero width,

$$\boxed{\operatorname{P}(X=a)=0}$$

for every continuous random variable. Consequently,

$$\operatorname{P}(X<a)=\operatorname{P}(X\leq a)$$

and similarly

$$\operatorname{P}(X>a)=\operatorname{P}(X\geq a).$$

This differs fundamentally from a discrete distribution, where $\operatorname{P}(X=a)$ may be positive.

Normal distribution calculators usually offer:

• normal CDF or normal cumulative probability for an area between bounds;
• inverse normal for a boundary corresponding to a given cumulative area.

Always identify the required region before entering values. For an upper tail, either use an upper bound of $+\infty$ or calculate a complement.

Worked example 2: a lower-tail probability

The lifetime $L$ hours of a component is modelled by

$$L\sim\operatorname{N}(1200,150^2).$$

Find the probability that a component lasts fewer than $1000$ hours.

The required area is to the left of $1000$:

$$\operatorname{P}(L<1000).$$

Using normal CDF with lower bound $-\infty$, upper bound $1000$, mean $1200$ and standard deviation $150$ gives

$$\operatorname{P}(L<1000)=0.091211\ldots$$

so

$$\boxed{\operatorname{P}(L<1000)\approx0.0912}.$$

The answer is below $0.5$, as expected because $1000$ is below the mean.

Worked example 3: probability between two values

The heights $H$ cm of plants are modelled by

$$H\sim\operatorname{N}(42,5^2).$$

Find $\operatorname{P}(38<H<50)$.

Use lower bound $38$, upper bound $50$, mean $42$ and standard deviation $5$:

$$\begin{aligned} \operatorname{P}(38<H<50) &=0.733313\ldots\\ &\approx\boxed{0.733}. \end{aligned}$$

The interval extends $0.8$ standard deviations below the mean and $1.6$ standard deviations above it, so an answer greater than $0.5$ is reasonable.

Worked example 4: an upper-tail probability

Scores $S$ on a test are modelled by

$$S\sim\operatorname{N}(64,12^2).$$

Find the probability that a randomly selected score exceeds $80$.

The complement is often easiest:

$$\begin{aligned} \operatorname{P}(S>80) &=1-\operatorname{P}(S\leq80)\\ &=1-0.908788\ldots\\ &=0.091211\ldots \end{aligned}$$

Therefore

$$\boxed{\operatorname{P}(S>80)\approx0.0912}.$$

Using symmetry

The curve is symmetric about $\mu$, so

$$\operatorname{P}(X<\mu-a)=\operatorname{P}(X>\mu+a).$$

Also,

$$\operatorname{P}(X<\mu)=\operatorname{P}(X>\mu)=\frac12.$$

For example, if $X\sim\operatorname{N}(30,4^2)$, then

$$\operatorname{P}(X<23)=\operatorname{P}(X>37).$$

Self-check 2

Let $X\sim\operatorname{N}(70,8^2)$. Find:

1. $\operatorname{P}(X<62)$;
2. $\operatorname{P}(62<X<78)$;
3. $\operatorname{P}(X>70)$.

Answer

The values $62$ and $78$ are one standard deviation below and above the mean.

$$\begin{aligned} \operatorname{P}(X<62)&=\boxed{0.1587},\\ \operatorname{P}(62<X<78)&=\boxed{0.6827},\\ \operatorname{P}(X>70)&=\boxed{0.5}. \end{aligned}$$

Standardising with a z-score

Any normal random variable can be transformed to the standard normal distribution

$$Z\sim\operatorname{N}(0,1)$$

using

$$\boxed{Z=\frac{X-\mu}{\sigma}}.$$

For an observed value $x$, its z-score is

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

It measures signed distance from the mean in standard deviations:

• $z=0$ means the value equals the mean;
• $z=1.5$ means it is $1.5$ standard deviations above the mean;
• $z=-2$ means it is $2$ standard deviations below the mean.

Standardising subtracts the location $\mu$, then divides by the scale $\sigma$. The direction of an inequality is unchanged because $\sigma>0$.

Worked example 5: standardising a boundary

Let $X\sim\operatorname{N}(40,6^2)$. Find $\operatorname{P}(X<49)$ by standardising.

Convert the boundary to a z-score:

$$z=\frac{49-40}{6}=1.5.$$

Therefore

$$\begin{aligned} \operatorname{P}(X<49) &=\operatorname{P}\left(Z<\frac{49-40}{6}\right)\\ &=\operatorname{P}(Z<1.5)\\ &=0.933192\ldots \end{aligned}$$

and hence

$$\boxed{\operatorname{P}(X<49)\approx0.9332}.$$

Modern calculators can obtain the same result directly from $X\sim\operatorname{N}(40,6^2)$, but z-scores expose the reasoning and are essential in parameter problems.

Worked example 6: comparing different scales

Amira scores $72$ on a test with mean $60$ and standard deviation $8$. Ben scores $84$ on a different test with mean $70$ and standard deviation $10$. Who performed better relative to their group?

Calculate both z-scores:

$$z_A=\frac{72-60}{8}=1.5,$$ $$z_B=\frac{84-70}{10}=1.4.$$

Since $1.5>1.4$, Amira’s score lies further above her group’s mean. Therefore

$$\boxed{\text{Amira performed better relative to her group.}}$$

This comparison uses the normal models, not the raw scores alone.

Variance and standard deviation

In $X\sim\operatorname{N}(\mu,\sigma^2)$, the denominator of the z-score is $\sigma$, not $\sigma^2$. If the distribution is $\operatorname{N}(40,36)$, divide by $6$, not $36$.

Self-check 3

For $Y\sim\operatorname{N}(25,4^2)$:

1. find the z-score corresponding to $y=19$;
2. use symmetry to find $\operatorname{P}(Y>31)$ from $\operatorname{P}(Y<19)$.

Answer

$$z=\frac{19-25}{4}=-1.5.$$

The values $19$ and $31$ are equally far from the mean, so

$$\operatorname{P}(Y>31)=\operatorname{P}(Y<19)=\boxed{0.0668}$$

to four decimal places.

Inverse normal problems

An inverse normal calculation starts with an area and finds the corresponding boundary. Most calculators expect the cumulative area to the left of the boundary.

If

$$\operatorname{P}(X<a)=p,$$

enter left-tail area $p$. If

$$\operatorname{P}(X>a)=p,$$

enter left-tail area $1-p$.

Worked example 7: finding a percentile

The journey time $T$ minutes is modelled by

$$T\sim\operatorname{N}(35,6^2).$$

Find the time below which $90\%$ of journeys fall.

We need $t$ such that

$$\operatorname{P}(T<t)=0.90.$$

Using inverse normal with area $0.90$, mean $35$ and standard deviation $6$ gives

$$t=42.689\ldots$$

Therefore

$$\boxed{t\approx42.7\text{ minutes}}.$$

Worked example 8: an upper-tail cutoff

The diameter $D$ mm of a manufactured part is modelled by

$$D\sim\operatorname{N}(20,0.12^2).$$

The largest $2\%$ are rejected. Find the minimum diameter that is rejected.

If the cutoff is $d$, then

$$\operatorname{P}(D>d)=0.02.$$

An inverse normal function uses the area to the left:

$$\operatorname{P}(D<d)=1-0.02=0.98.$$

Using area $0.98$, mean $20$ and standard deviation $0.12$ gives

$$d=20.2464\ldots$$

so the cutoff is

$$\boxed{20.25\text{ mm approximately}}.$$

Worked example 9: a central interval

Let $X\sim\operatorname{N}(100,15^2)$. Find $a$ such that

$$\operatorname{P}(100-a<X<100+a)=0.95.$$

The omitted probability is

$$1-0.95=0.05.$$

Symmetry puts $0.025$ in each tail. The upper boundary therefore has cumulative area

$$1-0.025=0.975.$$

The $0.975$ standard normal quantile is $1.95996\ldots$, so

$$a=1.95996\ldots\times15=29.399\ldots$$

and

$$\boxed{a\approx29.4}.$$

Central area is not a tail area

A central area of $0.90$ leaves $0.10$ outside, split equally as $0.05$ in each tail. Its upper boundary therefore corresponds to cumulative area $0.95$, not $0.90$.

Self-check 4

Weights $W$ kg are modelled by $W\sim\operatorname{N}(75,11^2)$. Find:

1. the 25th percentile;
2. the cutoff exceeded by the heaviest $10\%$.

Answer

For the 25th percentile, use left-tail area $0.25$:

$$W=75+(-0.67449\ldots)(11)=\boxed{67.6\text{ kg approximately}}.$$

The heaviest $10\%$ lie above the cutoff, so its left-tail area is $0.90$:

$$W=75+(1.28155\ldots)(11)=\boxed{89.1\text{ kg approximately}}.$$

Finding unknown parameters

Probability statements can determine $\mu$ or $\sigma$. Translate each statement into a standard normal quantile, then use

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

Do not round the quantile early. Retain several decimal places until the final answer.

Worked example 10: finding the mean

Suppose $X\sim\operatorname{N}(\mu,5^2)$ and

$$\operatorname{P}(X<27)=0.80.$$

Find $\mu$.

The standard normal value with cumulative probability $0.80$ is

$$z=0.841621\ldots$$

Therefore

$$\frac{27-\mu}{5}=0.841621\ldots$$

so

$$\begin{aligned} 27-\mu&=4.20810\ldots,\\ \mu&=22.7918\ldots \end{aligned}$$

Hence

$$\boxed{\mu\approx22.8}.$$

The result is sensible: $27$ has $80\%$ of the distribution below it, so it must lie above the mean.

Worked example 11: finding both parameters

Let $X\sim\operatorname{N}(\mu,\sigma^2)$, where

$$\operatorname{P}(X<32)=0.10$$

and

$$\operatorname{P}(X<50)=0.90.$$

Find $\mu$ and $\sigma$.

The corresponding z-values are

$$z_{0.10}=-1.281552\ldots, \qquad z_{0.90}=1.281552\ldots$$

so

$$\frac{32-\mu}{\sigma}=-1.281552\ldots$$

and

$$\frac{50-\mu}{\sigma}=1.281552\ldots$$

The probabilities are symmetric, so $32$ and $50$ are equally far from the mean:

$$\mu=\frac{32+50}{2}=41.$$

Then

$$\frac{50-41}{\sigma}=1.281552\ldots,$$

giving

$$\sigma=\frac9{1.281552\ldots}=7.02273\ldots$$

Therefore

$$\boxed{\mu=41,\qquad \sigma\approx7.02}.$$

Worked example 12: a non-symmetric pair of conditions

Suppose $Y\sim\operatorname{N}(\mu,\sigma^2)$ and

$$\operatorname{P}(Y<24)=0.20, \qquad \operatorname{P}(Y<39)=0.70.$$

Find $\mu$ and $\sigma$.

From inverse normal,

$$z_{0.20}=-0.841621\ldots, \qquad z_{0.70}=0.524401\ldots$$

Hence

$$24=\mu-0.841621\sigma$$

and

$$39=\mu+0.524401\sigma.$$

Subtracting the first equation from the second gives

$$15=(0.524401+0.841621)\sigma.$$

Thus

$$\sigma=10.9808\ldots$$

and, substituting into the second equation,

$$\mu=39-0.524401(10.9808\ldots)=33.2419\ldots$$

Therefore

$$\boxed{\mu\approx33.2,\qquad\sigma\approx11.0}.$$

Self-check 5

Let $X\sim\operatorname{N}(\mu,4^2)$ and $\operatorname{P}(X>30)=0.05$. Find $\mu$.

Answer

The area below $30$ is $0.95$, whose z-value is $1.644854\ldots$. Therefore

$$\frac{30-\mu}{4}=1.644854\ldots$$

and

$$\boxed{\mu=23.4\text{ approximately}}.$$

When is a normal model appropriate?

A normal model may be reasonable when:

• the variable is continuous;
• the data are approximately symmetric and unimodal;
• frequencies taper smoothly on both sides of the centre;
• there are no strong outliers or physical boundaries close to the bulk of the data;
• contextual knowledge supports a stable population with many small sources of variation.

A histogram or other data presentation can reveal skewness, multiple peaks, gaps and outliers. Context still matters. A small sample can look roughly bell-shaped by chance, and some measurements cannot sensibly take the negative values that a normal model technically permits.

Normal does not mean ordinary

In statistics, normal names a particular mathematical distribution. It does not mean that observed values outside a chosen range are wrong, impossible or abnormal.

Normal approximation to the binomial

If

$$X\sim\operatorname{B}(n,p),$$

then

$$\operatorname{E}(X)=np, \qquad \operatorname{Var}(X)=np(1-p).$$

When $n$ is large and $p$ is not too close to $0$ or $1$, the binomial distribution is approximately normal:

$$\boxed{X\approx Y,\qquad Y\sim\operatorname{N}\bigl(np,np(1-p)\bigr)}.$$

A common practical condition is

$$np>5 \qquad\text{and}\qquad n(1-p)>5,$$

although the exact convention can vary. Both expected successes and expected failures must be sufficiently numerous. Check the convention required by your course.

Why continuity correction is necessary

The binomial variable $X$ takes integer values, whereas the normal variable $Y$ takes all real values. To make their probability regions align, each integer is represented by an interval of width $1$ centred on it.

For example, $X=12$ is represented by

$$11.5<Y<12.5.$$

This adjustment is the continuity correction.

Binomial event	Corrected normal event
$X=k$	$k-0.5<Y<k+0.5$
$X\leq k$	$Y<k+0.5$
$X<k$	$Y<k-0.5$
$X\geq k$	$Y>k-0.5$
$X>k$	$Y>k+0.5$
$a\leq X\leq b$	$a-0.5<Y<b+0.5$

The strict or inclusive signs do not affect continuous probabilities. The $0.5$ boundary shift does.

Worked example 13: a cumulative binomial probability

Let

$$X\sim\operatorname{B}(100,0.4).$$

Use a normal approximation to estimate $\operatorname{P}(X\leq45)$.

First check suitability:

$$np=40>5, \qquad n(1-p)=60>5.$$

The approximating distribution has

$$\mu=np=40$$

and

$$\sigma^2=np(1-p)=24, \qquad \sigma=\sqrt{24}.$$

Apply the continuity correction:

$$\operatorname{P}(X\leq45)\approx\operatorname{P}(Y<45.5),$$

where $Y\sim\operatorname{N}(40,24)$. Therefore

$$\begin{aligned} \operatorname{P}(X\leq45) &\approx\operatorname{P}\left(Z<\frac{45.5-40}{\sqrt{24}}\right)\\ &=\operatorname{P}(Z<1.12268\ldots)\\ &\approx\boxed{0.869}. \end{aligned}$$

Worked example 14: an interval

Each of $200$ seeds germinates independently with probability $0.7$. Let $X$ be the number that germinate. Estimate

$$\operatorname{P}(135\leq X\leq150)$$

using a normal approximation.

Here

$$X\sim\operatorname{B}(200,0.7).$$

The suitability conditions hold because

$$np=140>5, \qquad n(1-p)=60>5.$$

Use

$$Y\sim\operatorname{N}(140,42),$$

since

$$np(1-p)=200(0.7)(0.3)=42.$$

Continuity correction gives

$$\operatorname{P}(135\leq X\leq150) \approx \operatorname{P}(134.5<Y<150.5).$$

Using normal CDF with standard deviation $\sqrt{42}$,

$$\operatorname{P}(134.5<Y<150.5)=0.7494\ldots$$

so

$$\boxed{\operatorname{P}(135\leq X\leq150)\approx0.749}.$$

Worked example 15: exactly one value

For $X\sim\operatorname{B}(80,0.25)$, estimate $\operatorname{P}(X=18)$ using a normal approximation.

Both suitability values exceed $5$:

$$np=20, \qquad n(1-p)=60.$$

Use

$$Y\sim\operatorname{N}(20,15).$$

The single discrete value $18$ corresponds to the continuous interval from $17.5$ to $18.5$:

$$\begin{aligned} \operatorname{P}(X=18) &\approx\operatorname{P}(17.5<Y<18.5)\\ &\approx\boxed{0.0900}. \end{aligned}$$

Without continuity correction, the normal probability of exactly $18$ would be zero.

Self-check 6

Let $X\sim\operatorname{B}(120,0.3)$. Use a normal approximation to estimate $\operatorname{P}(X>40)$.

Answer

The conditions hold because $np=36$ and $n(1-p)=84$. Use

$$Y\sim\operatorname{N}(36,25.2).$$

Since $X>40$ means $X\geq41$, continuity correction gives

$$\operatorname{P}(X>40)\approx\operatorname{P}(Y>40.5).$$

Therefore

$$\operatorname{P}(X>40) \approx \operatorname{P}\left(Z>\frac{40.5-36}{\sqrt{25.2}}\right) =\boxed{0.185\text{ approximately}}.$$

Common misconceptions

• Treating $\sigma^2$ as $\sigma$. In $\operatorname{N}(\mu,\sigma^2)$, take the square root of the second parameter before using calculator fields that ask for standard deviation.
• Using curve height as probability. Probability is area under the curve over an interval.
• Thinking $\operatorname{P}(X=a)$ is positive. It is $0$ for a continuous random variable.
• Entering an upper-tail area into inverse normal. Convert it to the left-tail area first unless the calculator explicitly accepts a tail choice.
• Splitting a central area incorrectly. Divide the excluded area equally between the two tails.
• Assuming every symmetric data set is normal. A normal model must also be unimodal, smoothly tapering and contextually reasonable.
• Approximating a binomial without checking conditions. Verify both $np$ and $n(1-p)$.
• Forgetting continuity correction. Translate the integer event before using the continuous model.

Mixed self-check

1. $X\sim\operatorname{N}(52,7^2)$. Find $\operatorname{P}(45<X<60)$.
2. For the same distribution, find $k$ if $\operatorname{P}(X>k)=0.15$.
3. $Y\sim\operatorname{N}(\mu,3^2)$ and $\operatorname{P}(Y<18)=0.25$. Find $\mu$.
4. $B\sim\operatorname{B}(250,0.6)$. State an approximating normal distribution and the corrected event for $\operatorname{P}(B\geq160)$.

Answers

1. Using normal CDF,

$$\operatorname{P}(45<X<60)=\boxed{0.715\text{ approximately}}.$$

2. The area to the left of $k$ is $0.85$. Thus

$$k=52+z_{0.85}(7) =52+(1.03643\ldots)(7) =\boxed{59.3\text{ approximately}}.$$

3. Since $z_{0.25}=-0.67449\ldots$,

$$\frac{18-\mu}{3}=-0.67449\ldots$$

so

$$\boxed{\mu\approx20.0}.$$

4. The mean and variance are

$$np=150, \qquad np(1-p)=60.$$

Therefore

$$\boxed{Y\sim\operatorname{N}(150,60)}$$

is suitable, and

$$\boxed{\operatorname{P}(B\geq160)\approx\operatorname{P}(Y>159.5)}.$$

Exam checklist

Before finishing a normal distribution problem, check that you have:

1. identified $\mu$, $\sigma^2$ and $\sigma$ correctly;
2. translated the words into a probability region;
3. used a complement or symmetry correctly where needed;
4. converted an upper-tail area before inverse normal;
5. checked whether the numerical answer is plausible;
6. checked approximation conditions and applied continuity correction for a binomial variable;
7. stated probabilities between $0$ and $1$ and rounded only at the end.

Next steps

• Review choosing a distribution when deciding between statistical models.
• Apply normal probabilities to significance testing in normal hypothesis tests.
• Revisit hypothesis testing language before interpreting critical regions and p-values.
• Strengthen modelling judgements with probability modelling and outliers and cleaning data.