Topics

The normal distribution

Covers AS91586

What it models

The normal distribution models measurements that cluster symmetrically around a mean, with values becoming rarer the further they are from it.
Write it as:

$X \sim N(\mu, \sigma^2)$

where $\mu$ is the mean and $\sigma$ the standard deviation. Note the second parameter is conventionally the variance.
Typical situations: heights and masses of people, animals or produce; measurement errors; manufactured dimensions; exam scores across a large cohort.

The shape

Symmetric and bell-shaped, centred on $\mu$ , with $\text{mean} = \text{median} = \text{mode}$ .
$\sigma$ controls the width: a larger $\sigma$ gives a flatter, wider curve; a smaller $\sigma$ a taller, narrower one. The total area stays 1 either way.
The tails extend forever in both directions but the area beyond three standard deviations is tiny.

The 68–95–99.7 rule

Within	Approximate area
$\mu \pm 1\sigma$	68%
$\mu \pm 2\sigma$	95%
$\mu \pm 3\sigma$	99.7%

Use this as a sanity check on every normal calculation. If a value is 1.5 standard deviations above the mean and your calculator gives a tail probability of 0.4, something is wrong — the answer should be around 0.07.

Standardising: the $z$ -score

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

The $z$ -score says how many standard deviations a value sits above (positive) or below (negative) the mean.
Standardising converts any normal distribution to the standard normal, $Z \sim N(0, 1)$ , which is what the booklet's tables describe.
Even with a graphing calculator, compute the $z$ -score. It is the quantity you interpret in context — "this is 2.1 standard deviations above the mean, so it is unusual" — and it is often worth marks on its own.

Finding probabilities on a calculator

Use normalcdf with the lower bound, upper bound, $\mu$ and $\sigma$ :

Wanted	Calculator
$P(X < 70)$	normalcdf $(-10^{99},\ 70,\ \mu,\ \sigma)$
$P(X > 70)$	normalcdf $(70,\ 10^{99},\ \mu,\ \sigma)$
$P(65 < X < 75)$	normalcdf $(65,\ 75,\ \mu,\ \sigma)$

Use a very large number for an unbounded end, not the mean or zero.
Sketch the curve and shade the region before calculating. It takes five seconds and it stops you finding the complement of what you wanted.

The inverse normal — going backwards

When a question gives a probability and asks for the value, use invNorm.
Typical wordings and what they need:
- "the mass exceeded by 10% of items" → invNorm with area 0.90 to the left
- "the fastest 5% of times" → for times, fastest means smallest, so area 0.05 to the left
- "the middle 90%" → two values, at areas 0.05 and 0.95.
Most calculators take the area to the LEFT. Convert before entering, and write the conversion down.

Finding an unknown parameter

If a question gives a probability and one of $\mu$ $μ$ or $\sigma$ $σ$ , work backwards through the $z$ $z$ -score:
- Find $z$ from the probability using invNorm on the standard normal.
- Substitute into $z = \frac{x - \mu}{\sigma}$ and solve for the unknown.
With two unknowns you need two probability statements, giving two equations to solve simultaneously.

When a normal model is appropriate

Use it when the data are roughly symmetric and unimodal, and the quantity results from many small independent influences added together — which is what produces a bell shape.
Do not use it when:
- the data are clearly skewed — incomes, house prices, waiting times, rainfall
- the variable is bounded near the mean, so the tails would fall outside possible values
- the data are discrete with few values, or bimodal (often a sign of two mixed populations).
A quick check: if $\mu - 3\sigma$ is impossible — a negative time, a negative mass — the normal model fits badly at the low end and you should say so.

Tips and Tricks

Sketch and shade before you calculate. Almost every wrong normal answer is the complement of the right one, and a sketch prevents it.

Compute and quote the $z$ -score even when the calculator does not need it. It is the interpretable quantity, and "1.8 standard deviations above the mean" is what earns the contextual mark.

Check against 68–95–99.7. A $z$ near 2 should give a tail of about 0.025; a $z$ near 1 about 0.16.

For invNorm, convert to the area on the left first, and write it down. "Top 15% means area 0.85 to the left" is a line worth having on the page.

Do not round $\mu$ or $\sigma$ before standardising. Early rounding is penalised, and $z$ -scores are sensitive to it.

When asked whether a normal model is appropriate, check the symmetry of the context and whether $\mu - 3\sigma$ is a possible value. If it is negative for a quantity that cannot be negative, say so.

Worked ExampleNormal probabilities, inverse normal, and judging the model

A packhouse grades kiwifruit by mass. The masses of a variety, $X$ grams, are normally distributed with mean 95 g and standard deviation 12 g.

(a) Find the probability a randomly chosen fruit has mass over 110 g. (b) Find the probability a fruit's mass is between 85 g and 105 g. (c) The heaviest 8% are sold as premium. Find the minimum mass for premium grade. (d) Fruit lighter than 70 g is rejected. Find the expected number rejected in a bin of 400 fruit. (e) The packhouse manager says the same normal model, with the mean adjusted, will describe a new variety whose fruit are much smaller, averaging 32 g with a standard deviation of 12 g. Comment.

(a) Mass over 110 g

Step 1 — Standardise.

$z = \frac{110 - 95}{12} = \frac{15}{12} = 1.25$

Step 2 — Sketch and shade the upper tail beyond $z = 1.25$ .

Step 3 — Calculate.

$P(X > 110) = \text{normalcdf}(110,\ 10^{99},\ 95,\ 12) = 0.10565...$

Check against the rule: $z = 1.25$ is a little above 1, and the tail beyond $z=1$ is about 16%, so an answer near 11% is sensible ✓

(b) Between 85 g and 105 g

Step 1 — Standardise both ends.

$z_1 = \frac{85 - 95}{12} = -0.8333... \qquad z_2 = \frac{105 - 95}{12} = 0.8333...$

The two $z$ -scores are symmetric about zero, which is expected since 85 and 105 are each 10 g from the mean.

Step 2 — Calculate.

$P(85 < X < 105) = \text{normalcdf}(85,\ 105,\ 95,\ 12) = 0.59526...$

Check: the band is about $\pm 0.83\sigma$ , a bit narrower than $\pm 1\sigma$ (68%), so 60% is sensible ✓

(c) The premium threshold — inverse normal

Step 1 — Convert to an area on the left. The heaviest 8% lie in the upper tail, so the threshold has

$\text{area to the left} = 1 - 0.08 = 0.92$

Step 2 — Apply invNorm.

$x = \text{invNorm}(0.92,\ 95,\ 12) = 111.86...$

Step 3 — Check via the $z$ -score. $\text{invNorm}(0.92)$ on the standard normal gives $z = 1.4051$ , so

$x = 95 + 1.4051 \times 12 = 111.86 \ \checkmark$

Interpret: premium fruit are those more than about 1.4 standard deviations above the mean mass.

(d) Expected number rejected from 400

Step 1 — Standardise.

$z = \frac{70 - 95}{12} = -2.0833...$

Step 2 — Find the probability.

$P(X < 70) = \text{normalcdf}(-10^{99},\ 70,\ 95,\ 12) = 0.018615...$

Step 3 — Scale to the bin.

$400 \times 0.018615... = 7.446$

Note the structure: this combines a normal probability with a count. Strictly the number rejected is $\text{Bin}(400, 0.0186)$ , whose mean is $np = 7.45$ — the same answer, and worth recognising as a link between the two model families.

(e) The new variety

Step 1 — Apply the check on the lower tail.

$\mu - 3\sigma = 32 - 3(12) = 32 - 36 = -4 \text{ grams}$

Step 2 — Quantify how bad it is.

$P(X < 0) = \text{normalcdf}(-10^{99},\ 0,\ 32,\ 12) = 0.00383$

So the model claims about 0.4% of fruit have negative mass — small, but a clear sign the model does not fit the situation.

Step 3 — Identify the real problem, which is not just the impossible values.

Step 4 — Check that fixes it.

$\mu - 3\sigma = 32 - 12 = 20 \text{ g}$

Step 5 — Conclude.

The normal distribution

Covers AS91586

What it models

The normal distribution models measurements that cluster symmetrically around a mean, with values becoming rarer the further they are from it.
Write it as:

$X \sim N(\mu, \sigma^2)$

where $\mu$ is the mean and $\sigma$ the standard deviation. Note the second parameter is conventionally the variance.
Typical situations: heights and masses of people, animals or produce; measurement errors; manufactured dimensions; exam scores across a large cohort.

The shape

Symmetric and bell-shaped, centred on $\mu$ , with $\text{mean} = \text{median} = \text{mode}$ .
$\sigma$ controls the width: a larger $\sigma$ gives a flatter, wider curve; a smaller $\sigma$ a taller, narrower one. The total area stays 1 either way.
The tails extend forever in both directions but the area beyond three standard deviations is tiny.

The 68–95–99.7 rule

Within	Approximate area
$\mu \pm 1\sigma$	68%
$\mu \pm 2\sigma$	95%
$\mu \pm 3\sigma$	99.7%

Use this as a sanity check on every normal calculation. If a value is 1.5 standard deviations above the mean and your calculator gives a tail probability of 0.4, something is wrong — the answer should be around 0.07.

Standardising: the $z$ -score

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

The $z$ -score says how many standard deviations a value sits above (positive) or below (negative) the mean.
Standardising converts any normal distribution to the standard normal, $Z \sim N(0, 1)$ , which is what the booklet's tables describe.
Even with a graphing calculator, compute the $z$ -score. It is the quantity you interpret in context — "this is 2.1 standard deviations above the mean, so it is unusual" — and it is often worth marks on its own.

Finding probabilities on a calculator

Use normalcdf with the lower bound, upper bound, $\mu$ and $\sigma$ :

Wanted	Calculator
$P(X < 70)$	normalcdf $(-10^{99},\ 70,\ \mu,\ \sigma)$
$P(X > 70)$	normalcdf $(70,\ 10^{99},\ \mu,\ \sigma)$
$P(65 < X < 75)$	normalcdf $(65,\ 75,\ \mu,\ \sigma)$

Use a very large number for an unbounded end, not the mean or zero.
Sketch the curve and shade the region before calculating. It takes five seconds and it stops you finding the complement of what you wanted.

The inverse normal — going backwards

When a question gives a probability and asks for the value, use invNorm.
Typical wordings and what they need:
- "the mass exceeded by 10% of items" → invNorm with area 0.90 to the left
- "the fastest 5% of times" → for times, fastest means smallest, so area 0.05 to the left
- "the middle 90%" → two values, at areas 0.05 and 0.95.
Most calculators take the area to the LEFT. Convert before entering, and write the conversion down.

Finding an unknown parameter

If a question gives a probability and one of $\mu$ $μ$ or $\sigma$ $σ$ , work backwards through the $z$ $z$ -score:
- Find $z$ from the probability using invNorm on the standard normal.
- Substitute into $z = \frac{x - \mu}{\sigma}$ and solve for the unknown.
With two unknowns you need two probability statements, giving two equations to solve simultaneously.

When a normal model is appropriate

Use it when the data are roughly symmetric and unimodal, and the quantity results from many small independent influences added together — which is what produces a bell shape.
Do not use it when:
- the data are clearly skewed — incomes, house prices, waiting times, rainfall
- the variable is bounded near the mean, so the tails would fall outside possible values
- the data are discrete with few values, or bimodal (often a sign of two mixed populations).
A quick check: if $\mu - 3\sigma$ is impossible — a negative time, a negative mass — the normal model fits badly at the low end and you should say so.

Tips and Tricks

Sketch and shade before you calculate. Almost every wrong normal answer is the complement of the right one, and a sketch prevents it.

Compute and quote the $z$ -score even when the calculator does not need it. It is the interpretable quantity, and "1.8 standard deviations above the mean" is what earns the contextual mark.

Check against 68–95–99.7. A $z$ near 2 should give a tail of about 0.025; a $z$ near 1 about 0.16.

For invNorm, convert to the area on the left first, and write it down. "Top 15% means area 0.85 to the left" is a line worth having on the page.

Do not round $\mu$ or $\sigma$ before standardising. Early rounding is penalised, and $z$ -scores are sensitive to it.

Worked ExampleNormal probabilities, inverse normal, and judging the model

A packhouse grades kiwifruit by mass. The masses of a variety, $X$ grams, are normally distributed with mean 95 g and standard deviation 12 g.

(a) Mass over 110 g

Step 1 — Standardise.

$z = \frac{110 - 95}{12} = \frac{15}{12} = 1.25$

Step 2 — Sketch and shade the upper tail beyond $z = 1.25$ .

Step 3 — Calculate.

$P(X > 110) = \text{normalcdf}(110,\ 10^{99},\ 95,\ 12) = 0.10565...$

Check against the rule: $z = 1.25$ is a little above 1, and the tail beyond $z=1$ is about 16%, so an answer near 11% is sensible ✓

(b) Between 85 g and 105 g

Step 1 — Standardise both ends.

$z_1 = \frac{85 - 95}{12} = -0.8333... \qquad z_2 = \frac{105 - 95}{12} = 0.8333...$

The two $z$ -scores are symmetric about zero, which is expected since 85 and 105 are each 10 g from the mean.

Step 2 — Calculate.

$P(85 < X < 105) = \text{normalcdf}(85,\ 105,\ 95,\ 12) = 0.59526...$

Check: the band is about $\pm 0.83\sigma$ , a bit narrower than $\pm 1\sigma$ (68%), so 60% is sensible ✓

(c) The premium threshold — inverse normal

Step 1 — Convert to an area on the left. The heaviest 8% lie in the upper tail, so the threshold has

$\text{area to the left} = 1 - 0.08 = 0.92$

Step 2 — Apply invNorm.

$x = \text{invNorm}(0.92,\ 95,\ 12) = 111.86...$

Step 3 — Check via the $z$ -score. $\text{invNorm}(0.92)$ on the standard normal gives $z = 1.4051$ , so

$x = 95 + 1.4051 \times 12 = 111.86 \ \checkmark$

Interpret: premium fruit are those more than about 1.4 standard deviations above the mean mass.

(d) Expected number rejected from 400

Step 1 — Standardise.

$z = \frac{70 - 95}{12} = -2.0833...$

Step 2 — Find the probability.

$P(X < 70) = \text{normalcdf}(-10^{99},\ 70,\ 95,\ 12) = 0.018615...$

Step 3 — Scale to the bin.

$400 \times 0.018615... = 7.446$

(e) The new variety

Step 1 — Apply the check on the lower tail.

$\mu - 3\sigma = 32 - 3(12) = 32 - 36 = -4 \text{ grams}$

Step 2 — Quantify how bad it is.

$P(X < 0) = \text{normalcdf}(-10^{99},\ 0,\ 32,\ 12) = 0.00383$

So the model claims about 0.4% of fruit have negative mass — small, but a clear sign the model does not fit the situation.

Step 3 — Identify the real problem, which is not just the impossible values.

Step 4 — Check that fixes it.

$\mu - 3\sigma = 32 - 12 = 20 \text{ g}$

Step 5 — Conclude.

The normal distribution

What it models

The shape

The 68–95–99.7 rule

Standardising: the zzz-score

Finding probabilities on a calculator

The inverse normal — going backwards

Finding an unknown parameter

When a normal model is appropriate

(a) Mass over 110 g

(b) Between 85 g and 105 g

(c) The premium threshold — inverse normal

(d) Expected number rejected from 400

(e) The new variety

The normal distribution

What it models

The shape

The 68–95–99.7 rule

Standardising: the zzz-score

Finding probabilities on a calculator

The inverse normal — going backwards

Finding an unknown parameter

When a normal model is appropriate

(a) Mass over 110 g

(b) Between 85 g and 105 g

(c) The premium threshold — inverse normal

(d) Expected number rejected from 400

(e) The new variety

Standardising: the $z$ -score

Standardising: the $z$ -score