Inverse Normal and Parameters (AA SL)
Some normal-distribution questions run backwards: instead of "find this probability", they give you a probability and ask for the boundary value, or for an unknown mean or standard deviation. These use the inverse normal function and a little algebra rather than a direct calculation. This page covers both, with worked examples and the mistakes that lose marks. It's part of the broader Random Variables & Distributions topic.
13 questions on this sub-topic.
Notation and standardising
Covered under IB syllabus reference SL4.9: the normal distribution and curve, its properties, and inverse normal calculations, found using technology.
Notation
\(X\sim N(\mu,\sigma^2)\)
\(\mu\) is the mean and \(\sigma^2\) is the variance. This is standard notation rather than a formula, so it isn't in the formula booklet.
Standardising
\(z = \dfrac{x-\mu}{\sigma}\)
Not in the formula booklet, but essential when \(\mu\) or \(\sigma\) is unknown - rearrange this equation once you know \(z\) from the inverse normal function.
Need the full syllabus wording and formula-booklet reference table? See Random Variables & Distributions.
Worked examples
\(X \sim N(70, 6^2)\).
Find the value \(k\) such that \(P(X < k) = 0.9\).
Worked solution
Need \(k\) with \(P(X
\(k=70+1.2816(6)\approx 77.7.\) A1
On the GDC: \(\texttt{invNorm(0.9, 70, 6)}\approx 77.7.\) Enter the left area, mean, then SD.
Lifetimes of bulbs are \(X\sim N(800, \sigma^2)\) hours. \(P(X < 740) = 0.10\).
Find \(\sigma\).
Worked solution
The z-score for a left tail of 0.10 is \(z=-1.282.\) A1
Standardising gives \(-1.282=\dfrac{740-800}{\sigma}=\dfrac{-60}{\sigma}.\) M1
\(\sigma=\dfrac{60}{1.282}\approx 46.8\) hours. A1
On the GDC: get the z-score with \(\texttt{invNorm(0.10,0,1)}\approx -1.282,\) then solve for \(\sigma\) algebraically - the GDC can't isolate an unknown parameter for you directly.
\(X\sim N(\mu, 4^2)\) and \(P(X<20)=0.7.\)
Find \(\mu.\)
Worked solution
\(z\) for \(0.7\) is \(\approx 0.5244.\) M1 A1
\(0.5244 = \dfrac{20 - \mu}{4}\) M1 \(\mu = 20 - 4(0.5244)\approx 17.9.\) A1
Common mistakes
- Entering the wrong area into invNorm. The inverse normal function always wants the area to the left of the boundary - if the question gives a right-tail probability (e.g. "top 15%"), convert it to a left area first by subtracting from 1.
- Trying to run invNorm with two unknown parameters. The GDC's inverse normal function needs numerical \(\mu\) and \(\sigma\) - when one of them is the unknown you're solving for, find \(z\) from the standard normal first, then substitute into \(z=\frac{x-\mu}{\sigma}\) by hand.
- Not rounding a cutoff value the right way. When a boundary represents a real quantity like a mark or a measurement, check whether the context needs rounding up or down - "the minimum mark for a distinction" rounds up, not to the nearest whole number.
Ready to practise properly?
11 inverse normal and parameter questions, marked instantly like the real exam.
Quick answers
What is an inverse normal calculation?
It works backwards from a known probability to find the boundary value \(k\), using the GDC's inverse normal function with the left-tail area as the input.
How do you find an unknown mean or standard deviation of a normal distribution?
Convert the given probability into a \(z\)-value with the inverse normal function on the standard normal, then substitute into \(z=\dfrac{x-\mu}{\sigma}\) and solve algebraically for the missing parameter.