Normal probabilities and inverse-normal values | Year 12 | Mathematical Methods

01 / LEARN

Understand the idea.

Areas in a normal model

Write $Φ (z) = P (Z \leq z)$ for standard normal lower-tail area. For a middle interval, subtract the lower cumulative area from the upper one. For an upper tail, use $1 - Φ (z)$ . A sketch helps keep the required region clear.

Normal areas are evaluated numerically. State the model, bounds, mean and standard deviation so the calculation can be reproduced. Keep unrounded z-scores internally, particularly when calculating small tail probabilities where rounding can have a larger relative effect.

P (a < X < b) = Φ (\frac{b - μ}{σ}) - Φ (\frac{a - μ}{σ})

Turn a percentage into a boundary

An inverse-normal function reverses an area calculation: it finds a boundary from the cumulative probability to its left. The highest $10 %$ begins at a boundary with $90 %$ below it, so its lower-tail input is $0.90$ .

After finding a standard normal quantile $z$ , return to the measurement scale with $x = μ + z σ$ . If a parameter is unknown, solve this equation instead. Check the resulting standard deviation is positive and the boundary lies on the expected side of the mean.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

An interval

$X \sim N (100, 64)$ . Find P(92<X<116).

Worked solution

1
Read the standard deviation and convert both bounds.
$\begin{aligned} σ & = \sqrt{64} = 8 \\ z_{lower} & = \frac{92 - 100}{8} = - 1 \\ z_{upper} & = \frac{116 - 100}{8} = 2 \end{aligned}$
Both boundaries must use the same centre and scale.
2
Subtract lower-tail areas.
$P (92 < X < 116) = P (- 1 < Z < 2) = Φ (2) - Φ (- 1)$
The smaller cumulative region is contained inside the larger one.
3
Evaluate without rounding intermediate boundaries.
$P (92 < X < 116) \approx 0.9772499 - 0.1586553 \approx 0.8186$
The result measures only the area between the two specified bounds.

Answer

0.8186.

EXAMPLE 02

The highest tenth

A normal model has mean 52 and SD 4. Find the cut-off for the highest 10%.

Worked solution

1
Convert the upper-tail target to a lower-tail target.
$P (X > c) = 0.10 \Rightarrow P (X \leq c) = 0.90$
The boundary must leave nine tenths of the continuous distribution below it.
2
Find the corresponding standard normal quantile.
$z = Φ^{- 1} (0.90) \approx 1.2815516$
The required percentile is above the mean, so its z-score is positive.
3
Restore the original scale.
$c = 52 + 4 Φ^{- 1} (0.90) \approx 57.1262$
The shift and scale convert the quantile to the original measurement units.

Answer

About 57.1262.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the standard normal 97.5th percentile.

Show answer & reasoning

1.9600.

The 97.5th percentile satisfies $Φ (z) = 0.975$ . Hence $z = Φ^{- 1} (0.975) \approx 1.9600$ , with $2.5 %$ remaining above it.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

$X \sim N (100, 64)$ . Find $P (X > 112)$ .
2 marks
A normal model has mean 50 and SD 6. Find its 2.5th percentile.
2 marks
Find the central 90% interval for a normal model with mean 52 and SD 4.
3 marks
A normal variable has SD 8 and $P (X < 80) = 0.90$ . Find its mean.
3 marks
For standard normal $Z$ , find $P (Z > 1 ∣ Z > 0)$ .
2 marks
A calculator returns -0.12 for an interval probability. What should you check?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$X \sim N (100, 64)$ . Find $P (X > 112)$ .
2 marks
$z = 1.5$ ; calculate $1 - Φ (1.5)$ .
0.0668.
A normal model has mean 50 and SD 6. Find its 2.5th percentile.
2 marks
$x = 50 + 6 Φ^{- 1} (0.025)$ .
38.2402.
Find the central 90% interval for a normal model with mean 52 and SD 4.
3 marks
There is 5% in each tail.
Use $52 \pm 4 Φ^{- 1} (0.95)$ .
$[45.4206, 58.5794]$ .
A normal variable has SD 8 and $P (X < 80) = 0.90$ . Find its mean.
3 marks
$\frac{80 - μ}{8} = Φ^{- 1} (0.90)$ .
Solve $μ = 80 - 8 Φ^{- 1} (0.90)$ .
69.7476.
For standard normal $Z$ , find $P (Z > 1 ∣ Z > 0)$ .
2 marks
Divide $P (Z > 1)$ by $P (Z > 0) = 0.5$ .
0.3173.
A calculator returns -0.12 for an interval probability. What should you check?
2 marks
Probabilities cannot be negative. Subtract the lower CDF from the upper CDF.
The calculation is invalid; check bound order and subtraction.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Areas in a normal model

Turn a percentage into a boundary

See the method in action.

An interval

Read the standard deviation and convert both bounds.

Subtract lower-tail areas.

Evaluate without rounding intermediate boundaries.

The highest tenth

Convert the upper-tail target to a lower-tail target.

Find the corresponding standard normal quantile.

Restore the original scale.

Try it yourself.

Check your understanding

Keep practising