IB Math AA SL Topic 4.12: z-values & inverse normal | Notes & Flashcards | Aimnova

IB Math AA SL — z-values & inverse normal

Topic 4.12 of IB Mathematics: Analysis and Approaches covers z-values & inverse normal, which is part of Unit 4: Statistics & Probability. Students explore key concepts including z-values, Inverse normal. A strong understanding of z-values & inverse normal is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in z-values & inverse normal

Key Idea: This topic links a value to its probability in a normal distribution: z-values measure how far a value sits from the mean, and the inverse normal turns a probability back into a value — a heavy Paper-2 GDC skill.

📏 z-values: standardising

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

$x$: the value you are standardising
$\mu$: the mean of the distribution
$\sigma$: the standard deviation
$z$: number of standard deviations from the mean (sign = side)

Positive z → above the mean; negative z → below. z is unitless (a count of standard deviations), so it lets you compare results from different distributions — the larger z is the better relative result. Standardising turns any X ~ N(μ, σ²) into the standard normal Z ~ N(0, 1).

🔁 Inverse normal: area → value

Important: invNorm(area, μ, σ) takes the lower / left-tail probability P(X < x) and returns the value x. Convert anything else first: • P(X > x) = p → use the left area 1 − p. • Middle proportion → split the leftover equally into two tails.

Tip: Find the standardised z = invNorm(p, 0, 1), then rearrange z = (x − μ)/σ: σ = (x − μ)/z or μ = x − zσ. You cannot feed the unknown σ straight into invNorm.

✏️ IB-style worked examples

IB-style question — find a z-value

The heights of seedlings are X ~ N(60, 8²) mm. Find the z-value of a seedling that is 72 mm tall.

Step by step:

Substitute into z = (x − μ)/σ.
$z = \frac{72 - 60}{8}$
Work it out.
$z = 1.5$

Final answer:

z = 1.5 — the seedling is 1.5 standard deviations above the mean.

IB-style question — compare across distributions

In Maths (mean 70, sd 5) Leah scores 80. In History (mean 60, sd 10) she scores 75. Relative to each class, which result is stronger?

Step by step:

Standardise each score.
$z_M = \frac{80-70}{5} = 2,\quad z_H = \frac{75-60}{10} = 1.5$
The larger z is further above its own mean.
$z_M > z_H$

Final answer:

Maths is the stronger result (z = 2 beats z = 1.5), even though 75 < 80 in raw marks.

IB-style question — inverse normal (left tail)

The mass of an apple is X ~ N(50, 10²) g. Find the value x such that P(X < x) = 0.9.

Step by step:

P(X < x) is already a left area — use invNorm.
$x = \text{invNorm}(0.9,\ 50,\ 10)$
Read the GDC result.
$x \approx 62.8$

Final answer:

x ≈ 62.8 g.

IB-style question — 'greater than' probability

For the same apples X ~ N(50, 10²) g, find x such that P(X > x) = 0.05 (the heaviest 5%).

Step by step:

invNorm needs the LEFT area — convert.
$P(X < x) = 1 - 0.05 = 0.95$
Apply invNorm with 0.95.
$x = \text{invNorm}(0.95,\ 50,\ 10) \approx 66.4$

Final answer:

x ≈ 66.4 g.

IB-style question — find an unknown σ

X ~ N(100, σ²) and P(X < 118) = 0.8. Find the standard deviation σ.

Step by step:

Get the standardised z for a left area of 0.8.
$z = \text{invNorm}(0.8,\ 0,\ 1) \approx 0.8416$
Rearrange z = (x − μ)/σ to σ = (x − μ)/z.
$\sigma = \frac{118 - 100}{0.8416} \approx 21.4$

Final answer:

σ ≈ 21.4.

IB-style question — find an unknown μ

X ~ N(μ, 5²) and P(X < 20) = 0.1. Find the mean μ.

Step by step:

Get z for a left area of 0.1 (it is negative).
$z = \text{invNorm}(0.1,\ 0,\ 1) \approx -1.2816$
Rearrange to μ = x − zσ.
$\mu = 20 - (-1.2816)(5) \approx 26.4$

Final answer:

μ ≈ 26.4 — since 20 lies in the lower tail, the mean is above it.

Important: invNorm always works on the lower / left-tail probability. 'Top 10%' is NOT area 0.10 — it is P(X > x) = 0.10, so the left area is 1 − 0.10 = 0.90. Convert every 'greater than' or 'top %' to a left area before you type it in.

Tap each card to reveal the answer.

Exam Tips

z = (x − μ)/σ: positive is above the mean, negative below; z has no units.
Compare results from different distributions by z, never by raw scores.
invNorm(area, μ, σ) uses the LEFT-tail area — convert 'greater than' to 1 − p.
For an unknown σ or μ, get z = invNorm(p, 0, 1), then use z = (x − μ)/σ.
A left area below 0.5 returns a negative z — keep the sign in your rearrangement.

IB Math AA SL — z-values & inverse normal

Key concepts in z-values & inverse normal

Key Idea: This topic links a value to its probability in a normal distribution: z-values measure how far a value sits from the mean, and the inverse normal turns a probability back into a value — a heavy Paper-2 GDC skill.

📏 z-values: standardising

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

$x$: the value you are standardising
$\mu$: the mean of the distribution
$\sigma$: the standard deviation
$z$: number of standard deviations from the mean (sign = side)

Positive z → above the mean; negative z → below. z is unitless (a count of standard deviations), so it lets you compare results from different distributions — the larger z is the better relative result. Standardising turns any X ~ N(μ, σ²) into the standard normal Z ~ N(0, 1).

🔁 Inverse normal: area → value

Important: invNorm(area, μ, σ) takes the lower / left-tail probability P(X < x) and returns the value x. Convert anything else first: • P(X > x) = p → use the left area 1 − p. • Middle proportion → split the leftover equally into two tails.

Tip: Find the standardised z = invNorm(p, 0, 1), then rearrange z = (x − μ)/σ: σ = (x − μ)/z or μ = x − zσ. You cannot feed the unknown σ straight into invNorm.

✏️ IB-style worked examples

IB-style question — find a z-value

The heights of seedlings are X ~ N(60, 8²) mm. Find the z-value of a seedling that is 72 mm tall.

Step by step:

Substitute into z = (x − μ)/σ.
$z = \frac{72 - 60}{8}$
Work it out.
$z = 1.5$

Final answer:

z = 1.5 — the seedling is 1.5 standard deviations above the mean.

IB-style question — compare across distributions

In Maths (mean 70, sd 5) Leah scores 80. In History (mean 60, sd 10) she scores 75. Relative to each class, which result is stronger?

Step by step:

Standardise each score.
$z_M = \frac{80-70}{5} = 2,\quad z_H = \frac{75-60}{10} = 1.5$
The larger z is further above its own mean.
$z_M > z_H$

Final answer:

Maths is the stronger result (z = 2 beats z = 1.5), even though 75 < 80 in raw marks.

IB-style question — inverse normal (left tail)

The mass of an apple is X ~ N(50, 10²) g. Find the value x such that P(X < x) = 0.9.

Step by step:

P(X < x) is already a left area — use invNorm.
$x = \text{invNorm}(0.9,\ 50,\ 10)$
Read the GDC result.
$x \approx 62.8$

Final answer:

x ≈ 62.8 g.

IB-style question — 'greater than' probability

For the same apples X ~ N(50, 10²) g, find x such that P(X > x) = 0.05 (the heaviest 5%).

Step by step:

invNorm needs the LEFT area — convert.
$P(X < x) = 1 - 0.05 = 0.95$
Apply invNorm with 0.95.
$x = \text{invNorm}(0.95,\ 50,\ 10) \approx 66.4$

Final answer:

x ≈ 66.4 g.

IB-style question — find an unknown σ

X ~ N(100, σ²) and P(X < 118) = 0.8. Find the standard deviation σ.

Step by step:

Get the standardised z for a left area of 0.8.
$z = \text{invNorm}(0.8,\ 0,\ 1) \approx 0.8416$
Rearrange z = (x − μ)/σ to σ = (x − μ)/z.
$\sigma = \frac{118 - 100}{0.8416} \approx 21.4$

Final answer:

σ ≈ 21.4.

IB-style question — find an unknown μ

X ~ N(μ, 5²) and P(X < 20) = 0.1. Find the mean μ.

Step by step:

Get z for a left area of 0.1 (it is negative).
$z = \text{invNorm}(0.1,\ 0,\ 1) \approx -1.2816$
Rearrange to μ = x − zσ.
$\mu = 20 - (-1.2816)(5) \approx 26.4$

Final answer:

μ ≈ 26.4 — since 20 lies in the lower tail, the mean is above it.

Important: invNorm always works on the lower / left-tail probability. 'Top 10%' is NOT area 0.10 — it is P(X > x) = 0.10, so the left area is 1 − 0.10 = 0.90. Convert every 'greater than' or 'top %' to a left area before you type it in.

Tap each card to reveal the answer.

Exam Tips

z = (x − μ)/σ: positive is above the mean, negative below; z has no units.
Compare results from different distributions by z, never by raw scores.
invNorm(area, μ, σ) uses the LEFT-tail area — convert 'greater than' to 1 − p.
For an unknown σ or μ, get z = invNorm(p, 0, 1), then use z = (x − μ)/σ.
A left area below 0.5 returns a negative z — keep the sign in your rearrangement.

IB Math AA SL — z-values & inverse normal

Key concepts in z-values & inverse normal

📏 z-values: standardising

🔁 Inverse normal: area → value

✏️ IB-style worked examples

IB-style question — find a z-value

IB-style question — compare across distributions

IB-style question — inverse normal (left tail)

IB-style question — 'greater than' probability

IB-style question — find an unknown σ

IB-style question — find an unknown μ

Exam Tips

What you'll learn in Topic 4.12

Study resources — 4.12 z-values & inverse normal

z-values

Inverse normal

Ready to study z-values & inverse normal?

Ready to practice?

IB Math AA SL — z-values & inverse normal

Key concepts in z-values & inverse normal

📏 z-values: standardising

🔁 Inverse normal: area → value

✏️ IB-style worked examples

IB-style question — find a z-value

IB-style question — compare across distributions

IB-style question — inverse normal (left tail)

IB-style question — 'greater than' probability

IB-style question — find an unknown σ

IB-style question — find an unknown μ

Exam Tips

What you'll learn in Topic 4.12

Study resources — 4.12 z-values & inverse normal

z-values

Inverse normal

Ready to study z-values & inverse normal?

Ready to practice?