How do you find the number of items meeting a condition?

First find the proportion with the normal CDF, then multiply by the population size. For example, if P(X less than 170) = 0.1587 and there are 2000 items, the expected number is 0.1587 times 2000, which is about 317. Round a count to a whole number.

How do you find a cutoff for the top 10 percent?

The top 10 percent lie above the 90th percentile, so the area below the cutoff is 0.90. Use the calculator's inverse normal with a lower area of 0.90 and the given mean and standard deviation. For X ~ N(60, 12^2) this gives invNorm(0.90, 60, 12) = 75.38.

How do you work backwards to find an unknown mean or standard deviation?

Turn the given probability into a z-value with the inverse normal, then substitute into z = (x - mu)/sigma and solve for the unknown. If 2.5 percent of values are below 90 and sigma is 4, then z = -1.96, so -1.96 = (90 - mu)/4, giving mu = 97.84.

How do you find both the mean and the standard deviation?

You need two probabilities. Convert each to a z-value and write z = (x - mu)/sigma as two linear equations in mu and sigma. Subtract the equations to eliminate mu and solve for sigma, then back-substitute to find mu.

When should you use the 68-95-99.7 rule instead of technology?

The empirical 68-95-99.7 rule only gives approximate proportions at whole numbers of standard deviations from the mean. When a problem asks for a probability at any other value, or for a precise cutoff, use technology (the normal CDF or inverse normal) for an exact answer.

How do you find the number of items meeting a condition?

First find the proportion with the normal CDF, then multiply by the population size. For example, if P(X less than 170) = 0.1587 and there are 2000 items, the expected number is 0.1587 times 2000, which is about 317. Round a count to a whole number.

How do you find a cutoff for the top 10 percent?

The top 10 percent lie above the 90th percentile, so the area below the cutoff is 0.90. Use the calculator's inverse normal with a lower area of 0.90 and the given mean and standard deviation. For X ~ N(60, 12^2) this gives invNorm(0.90, 60, 12) = 75.38.

How do you work backwards to find an unknown mean or standard deviation?

Turn the given probability into a z-value with the inverse normal, then substitute into z = (x - mu)/sigma and solve for the unknown. If 2.5 percent of values are below 90 and sigma is 4, then z = -1.96, so -1.96 = (90 - mu)/4, giving mu = 97.84.

How do you find both the mean and the standard deviation?

You need two probabilities. Convert each to a z-value and write z = (x - mu)/sigma as two linear equations in mu and sigma. Subtract the equations to eliminate mu and solve for sigma, then back-substitute to find mu.

When should you use the 68-95-99.7 rule instead of technology?

The empirical 68-95-99.7 rule only gives approximate proportions at whole numbers of standard deviations from the mean. When a problem asks for a probability at any other value, or for a precise cutoff, use technology (the normal CDF or inverse normal) for an exact answer.

Year 12 Methods (Unit 3 & 4) The normal distribution

Solving problems using the normal distribution

Q: How do you find a normal probability with technology?

Enter the mean and standard deviation into the calculator's normal CDF and give the lower and upper limits of the region. For P(X less than a) use a large negative lower limit; for P(X greater than a) use a large positive upper limit; for P(a less than X less than b) use a and b. The calculator returns the proportion (a number between 0 and 1).

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), solving problems using the normal distribution \(X\sim N(\mu,\sigma^{2})\) means applying technology to real contexts: a normal CDF gives a proportion, multiplying by the population gives a number, an inverse normal gives a cutoff or percentile, and standardising with \(z=\dfrac{x-\mu}{\sigma}\) lets you work backwards to an unknown \(\mu\) or \(\sigma\).

Many quantities — heights, weights, times, exam marks, manufacturing measurements — are modelled by a normal random variable \(X\sim N(\mu,\sigma^{2})\). Solving a problem means turning a real question into one of four standard tasks and answering it with QCAA-approved technology (distribution tables are not required).

A proportion is a probability read from the calculator's normal CDF: \(P(Xa)\) or \(P(anumber (or count) is that proportion multiplied by the population size. A cutoff or percentile — the mark for the top \(10\%\), a quartile — comes from the inverse normal, which reverses the CDF.

To find an unknown parameter, standardise: a value \(x\) is \(z=\dfrac{x-\mu}{\sigma}\) standard deviations from the mean. Turning a given probability into its \(z\)-value gives an equation you can solve for \(\mu\) or \(\sigma\); two probabilities give two equations, enough to find both.

Key idea. Proportion → normal CDF; number → proportion \(\times\) population; cutoff → inverse normal; unknown \(\mu\) or \(\sigma\) → standardise with \(z=\dfrac{x-\mu}{\sigma}\) and solve.

A normal CDF returns the shaded proportion \(P(X>a)\)

Inverse normal: the cutoff \(k\) for the top \(10\%\) satisfies \(P(X

A proportion from the normal CDF (a probability between \(0\) and \(1\)):

\[P(aa), \qquad X\sim N(\mu,\sigma^{2})\]

P (a < X < b)

The expected number meeting a condition (proportion times population \(n\)):

\[\text{number}=P(\text{condition})\times n\]

Standardising a value, and the cutoff for a given percentile:

\[z=\dfrac{x-\mu}{\sigma}, \qquad k=\text{invNorm}(p,\mu,\sigma) \ \text{ where } P(X

z = \frac{x - μ}{σ}

Both \(\mu\) and \(\sigma\). Two probabilities give two equations \(x_1=\mu+z_1\sigma\) and \(x_2=\mu+z_2\sigma\); subtract to eliminate \(\mu\) and solve for \(\sigma\), then back-substitute.

How to solve a normal distribution problem

Write the model. Identify \(X\sim N(\mu,\sigma^{2})\) from the context (the mean and the standard deviation).
Proportion? Use the normal CDF, taking care which tail is asked for: \(P(X>a)=1-P(X
Number? Multiply the proportion by the population size and round the count to a whole number.
Cutoff or percentile? Use the inverse normal. For the top \(10\%\) the lower area is \(0.90\); for the lower quartile it is \(0.25\).
Unknown \(\mu\) or \(\sigma\)? Convert the probability to a \(z\)-value, substitute into \(z=\dfrac{x-\mu}{\sigma}\) and solve. Two probabilities give two equations for both parameters.

Top 10% or bottom 25%? Always convert to a lower area for the inverse normal: the top \(10\%\) is a lower area of \(0.90\); the bottom quartile is a lower area of \(0.25\).

Example 1 — A proportion

Battery life is \(N(40,\,5^{2})\) hours. Find the probability that a battery lasts less than \(45\) hours.

Solution

Read the lower area from the normal CDF.

\(X\)	\(\sim\)	\(N(40,\,5^{2})\)
\(P(X<45)\)	\(=\)	\(\text{normalCDF}(-\infty,45,40,5)\)
	\(=\)	\(0.8413\)

So \(P(X<45)=0.8413\).

0.8413

Example 2 — Proportion to a number

Bag masses are \(N(175,\,5^{2})\) g. Of \(2000\) bags, how many weigh less than \(170\) g?

Solution

Find the proportion, then multiply by \(2000\).

\(P(X<170)\)	\(=\)	\(0.1587\)
number	\(=\)	\(0.1587\times 2000\)
	\(=\)	\(317\) bags

About \(317\) bags weigh under \(170\) g.

317

Example 3 — A cutoff (inverse normal)

Exam marks are \(N(60,\,12^{2})\). A prize goes to the top \(10\%\). Find the minimum mark.

Solution

The top \(10\%\) is a lower area of \(0.90\).

\(P(X	\(=\)	\(0.90\)
\(k\)	\(=\)	\(\text{invNorm}(0.90,60,12)\)
	\(=\)	\(75.38\)

The minimum mark is \(75.38\) (about \(76\)).

75.38

Example 4 — Work backwards for \(\mu\)

A machine has \(\sigma=4\) g, and \(2.5\%\) of bags weigh less than \(90\) g. Find the mean \(\mu\).

Solution

Standardise, then solve for \(\mu\).

\(z\)	\(=\)	\(\text{invNorm}(0.025)=-1.96\)
\(-1.96\)	\(=\)	\(\dfrac{90-\mu}{4}\)
\(\mu\)	\(=\)	\(97.84\)

The mean mass is \(97.84\) g.

μ = 97.84

Common pitfalls

Proportion versus number. A proportion is a probability between \(0\) and \(1\); a number of items is that proportion \(\times\) the population size. Read the question to see which is wanted, and round a count to a whole number.

Check the tail. \(P(X>a)=1-P(X

Solve for the parameter, not \(x\). When working backwards, rearrange \(z=\dfrac{x-\mu}{\sigma}\) for the unknown \(\mu\) or \(\sigma\). A left-tail \(z\) is negative — a dropped sign flips the answer.

Use technology, not the \(68\)–\(95\)–\(99.7\) rule, for exact values. The empirical rule only gives rough proportions at whole standard deviations; for any other value or a precise cutoff, use the normal CDF or inverse normal.

Frequently asked questions

How do you find a normal probability with technology?

Enter \(\mu\) and \(\sigma\) into the normal CDF with the region's lower and upper limits. For \(P(X>a)\) use a large upper limit; the result is a proportion between \(0\) and \(1\).

How do you find the number meeting a condition?

Find the proportion with the normal CDF, then multiply by the population size. E.g. \(0.1587\times 2000\approx 317\). Round a count to a whole number.

How do you find the cutoff for the top 10%?

The top \(10\%\) is a lower area of \(0.90\), so use \(\text{invNorm}(0.90,\mu,\sigma)\). For \(N(60,12^{2})\) this gives \(75.38\).

How do you work backwards to find mu or sigma?

Turn the probability into a \(z\)-value, substitute into \(z=\dfrac{x-\mu}{\sigma}\) and solve. E.g. \(-1.96=\dfrac{90-\mu}{4}\) gives \(\mu=97.84\).

How do you find both mu and sigma?

Use two probabilities to write two equations \(x=\mu+z\sigma\); subtract to eliminate \(\mu\) and solve for \(\sigma\), then back-substitute for \(\mu\).

When is the 68-95-99.7 rule not enough?

Whenever a value is not a whole number of standard deviations from the mean, or a precise cutoff is needed — then use the normal CDF or inverse normal for an exact answer.

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