IB Maths AA SL Topic 4 — Probability Distributions Paper 2 (GDC) ~10 min read

Finding Unknown Parameters

“5% of the bags weigh more than 510 g — find the standard deviation.” Sound familiar? This is a Paper 2 favourite. The trick: when a question gives you a probability and asks for the mean or SD, you have to work backwards using standardisation. Once you know the recipe, these become some of the highest-mark, most reliable questions to nail.

📘 What you need to know

An “unknown parameter” means the question hides μ or σ (or both) and asks you to find it from a probability.
The recipe always works the same way: get a z-value first, then substitute into Z = X − μσ, then solve.
Find z using Inverse Normal on N(0, 1), with the area to the LEFT of X.
Convert “above” probabilities first: P(X > a) = 0.10 → use 0.90 in Inverse Normal.
If both μ and σ are unknown, the question gives you two probabilities → set up two equations and solve simultaneously.
Sketch the curve for every question and mark which side X falls on — that’s how you know the sign of z.

The 3 scenarios you’ll meet

Every “find the unknown” question in IB AA SL falls into one of three buckets. Read the question, spot which one it is, then apply the matching method.

1 Find μ

Given: σ & one probability

Use the 3-step recipe. One equation, one unknown.

2 Find σ

Given: μ & one probability

Same 3-step recipe. Just a different unknown.

3 Find both μ & σ

Given: two probabilities

Get two equations. Solve simultaneously.

A quick way to spot which scenario you’re in: count how many things the question gives you. One probability → one of μ or σ is hidden. Two probabilities → both are hidden. The number of unknowns always matches the number of bits of info.

The recipe (works for every question)

Whether you’re finding μ, σ, or both, the actual steps barely change. Here’s the universal version:

The 3-step recipe

Sketch & identify. Draw a quick bell curve. Mark X on the axis. Note which side of the mean it sits on (this fixes the sign of z later).
Find z using Inverse Normal on the standard normal N(0, 1). Use the area to the left as your input. Convert “above” probabilities (1 − the given value) first.
Substitute into Z = (X − μ)/σ and solve for the unknown.

📍

Why use N(0, 1) in Step 2?

Standardising removes μ and σ from the picture, so the z-value depends only on the probability — nothing else. That’s why you always run Inverse Normal with mean = 0 and SD = 1 in this step. The actual μ and σ come back in when you substitute later.

Spotting the question type from the wording

Exam questions don’t always announce themselves clearly. Here’s how typical phrasings translate into maths:

If the question says…	It means…
“5% of items weigh more than 800 g”	P(X > 800) = 0.05 (top tail)
“The slowest 10% take more than…”	P(X > k) = 0.10 (top tail)
“The fastest 5% finish in less than…”	P(X < k) = 0.05 (bottom tail)
“P(X < 80) = 0.18”	use 0.18 directly in Inverse Normal
“The middle 95% lie between 60 and 80”	range is symmetric → 2.5% in each tail
“95% of bags weigh between A and B”	2.5% below A, 2.5% above B

The symmetric range trick

Here’s a slick shortcut for one specific kind of question. If a question gives you a range like “the middle 95% lie between 60 and 80” and the range is symmetric about the mean, then the mean is just the midpoint of the two values.

Symmetric range trick

μ = a + b2

So in that example, μ = (60 + 80)/2 = 70. You only need to find σ from there using one z-value.

🤔 Why does this trick work?

The normal distribution is symmetric about its mean. If the “middle 95%” lies between two values, those two values must be equally far from the mean — one above, one below. The midpoint between them is the mean itself.

This trick saves you from having to set up two equations. You get μ for free from the midpoint, then only have to find σ. Watch out though — it only works when the question explicitly says “middle X%” or makes it clear the range is symmetric. For unequal tail percentages, you need the full simultaneous-equations method.

When both μ and σ are unknown

This is the biggest version of the question. It looks scarier but follows the same recipe — just twice.

Sketch the curve. Mark both given X-values on it.
Find both z-values using Inverse Normal on N(0, 1).
Set up two equations using Z = (X − μ)/σ for each.
Subtract one equation from the other to eliminate μ — you’ll be left with one equation in σ. Solve.
Substitute σ back into either original equation to get μ.

🧠

“Subtract to kill μ”

When you write both equations as (X − μ) = z · σ, the −μ appears in both. Subtracting one from the other makes it disappear, leaving you a quick linear equation in σ alone. We’ll see this in WE 5.

Worked examples

WE 1

Find an unknown mean

The masses of bags of rice are normally distributed with standard deviation 8 g. The probability that a randomly chosen bag has mass less than 990 g is 0.10. Find the mean mass.

σ given, looking for μ. One probability → 3-step recipe.step 1 — sketch & identify P(X < 990) = 0.10 → 990 is in the lower tail, BELOW the mean. So z will be NEGATIVE.step 2 — find z Inverse Normal, area = 0.10, μ = 0, σ = 1 z = −1.2816…step 3 — substitute & solve −1.2816 = (990 − μ)/8 −1.2816 × 8 = 990 − μ −10.253 = 990 − μ → μ = 990 + 10.253 μ ≈ 1000 g sketch first ALWAYS — it’s the only reliable way to get the sign of z right!

WE 2

Find an unknown standard deviation

The lifetime of a light bulb is normally distributed with mean 1500 hours. It is known that 5% of bulbs last more than 1700 hours. Find the standard deviation.

μ given, looking for σ. “More than” → convert to area on left first.step 1 — sketch & identify P(X > 1700) = 0.05 → 1700 is ABOVE the mean. z is POSITIVE.step 2 — find z Convert: P(X < 1700) = 1 − 0.05 = 0.95 Inverse Normal, area = 0.95, μ = 0, σ = 1 z = 1.6449…step 3 — substitute & solve 1.6449 = (1700 − 1500)/σ = 200/σ σ = 200/1.6449 = 121.57… σ ≈ 122 hours “more than” → flip with 1 − 0.05 = 0.95 BEFORE Inverse Normal!

WE 3

Use the symmetric range trick

The heights of trees in a forest are normally distributed. The middle 95% of trees have heights between 14 m and 22 m. Find μ and σ.

“Middle 95%” → range is symmetric about μ. Mean is the midpoint!μ — midpoint trick μ = (14 + 22)/2 = 18 μ = 18 mσ — use one z Middle 95% → 2.5% in each tail. P(X < 22) = 0.975, so for X = 22: Inverse Normal, area = 0.975: z = 1.9599… 1.9599 = (22 − 18)/σ = 4/σ σ = 4/1.9599 = 2.040… σ ≈ 2.04 m “middle X%” = symmetric → mean is the midpoint, no equations needed for μ!

WE 4

Real exam-style: find σ from a percentile

The marks on a test are normally distributed with mean 60. The top 15% of students score above 75 marks. Find the standard deviation.

μ given. “Top 15%” → upper tail. Convert to left area first.step 1 — sketch & identify P(X > 75) = 0.15 → 75 is above mean. z > 0.step 2 — find z P(X < 75) = 1 − 0.15 = 0.85 Inverse Normal, area = 0.85: z = 1.0364…step 3 — substitute 1.0364 = (75 − 60)/σ = 15/σ σ = 15/1.0364 = 14.473… σ ≈ 14.5 marks 3 s.f. unless told otherwise — this is a Paper 2 staple!

WE 5

Find both μ and σ (simultaneous equations)

The weights of bags of flour are normally distributed. P(X < 950) = 0.05 and P(X > 1050) = 0.10. Find μ and σ.

Two unknowns → need two equations. Use Inverse Normal twice, then subtract to kill μ.step 1 — find both z-values For P(X < 950) = 0.05: Inverse Normal, area = 0.05: z₁ = −1.6449… For P(X > 1050) = 0.10 → P(X < 1050) = 0.90: Inverse Normal, area = 0.90: z₂ = 1.2816…step 2 — set up equations −1.6449 = (950 − μ)/σ ⇒ 950 − μ = −1.6449σ ① 1.2816 = (1050 − μ)/σ ⇒ 1050 − μ = 1.2816σ ② step 3 — subtract ① from ② (kills μ) (1050 − μ) − (950 − μ) = 1.2816σ − (−1.6449σ) 100 = 2.9265σ σ = 100/2.9265 = 34.17… σ ≈ 34.2step 4 — find μ Sub σ into ①: 950 − μ = −1.6449(34.17) = −56.20… μ = 950 + 56.20 = 1006.2… μ ≈ 1006 g always set up your equations as (X − μ) = zσ — it makes the subtraction step automatic!

💡 Top tips

Start with a sketch. Mark the value, shade the region, decide whether z is positive or negative. This single habit fixes most sign errors.
Always run Inverse Normal on N(0, 1) in Step 2 — set μ = 0 and σ = 1 in the calculator.
Convert “above” probabilities to area-on-left before Inverse Normal: 1 − (given probability).
For “middle X%” questions, use the midpoint trick: μ = (a + b)/2. You only need to find σ.
For both unknowns, set up equations as (X − μ) = zσ then subtract to eliminate μ. It’s much faster than expanding everything.
Keep z to 4 decimal places through your working. Rounding to 1 d.p. early ruins the final answer.
Sanity check the answer. Heights ≈ 1.7 m, ages 0–100, exam scores 0–100. If σ comes out as 800, something’s wrong.
Always include units in the final answer if the context has them.

⚠ Common mistakes

Wrong sign on z. If X is below the mean, z must be negative. Most sign errors come from skipping the sketch.
Using the upper tail directly in Inverse Normal. It only takes area to the LEFT — convert first.
Forgetting “middle 95%” gives 2.5% in each tail. It’s NOT 5% on one side.
Solving the simultaneous equations the slow way. Don’t expand and rearrange — just subtract to kill μ.
Using μ = 0, σ = 1 in the wrong step. That’s only for Inverse Normal in Step 2 — your real μ and σ go back into Step 3.
Treating the question as a Cdf problem. If it gives you a probability and asks for a value or parameter, it’s standardisation, not Cdf.
Rounding z too early. Always carry 4+ decimal places through to your final calculation.
Mis-reading the tail. “Below 80” with probability 0.18 is the LEFT tail. “Above 80” with the same number is the RIGHT tail. They give different z-signs.

🎉 You can now handle every “find the unknown” question that exams throw at you — finding μ, finding σ, or finding both at once. These questions are worth a lot of marks and once the recipe clicks, they’re very predictable. Practise a few from past papers and you’ll start spotting the pattern instantly.

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