Inverse Normal and Parameters (AA HL)
Most normal distribution questions give you an \(x\)-value and ask for a probability. This sub-topic runs the process backwards: you're given a probability and asked to find the \(x\)-value, the mean \(\mu\), or the standard deviation \(\sigma\) that produced it. It's part of the broader Probability Distributions topic.
11 questions on this sub-topic.
The key formula
Covered under IB syllabus reference SL4.9 - normal and inverse normal probabilities are always found using the GDC, not by hand. The standardizing formula below is what links the raw value \(x\) to the standard-normal \(z\)-value your calculator's inverse function actually returns.
Standardizing
\(z=\dfrac{x-\mu}{\sigma}\)
Rearrange this to solve for whichever quantity is unknown - \(x\), \(\mu\), or \(\sigma\) - once you have \(z\) from the GDC.
Inverse normal
Given a probability, find the corresponding \(x\)-value - always found using the GDC's inverse normal function, working from the area to the LEFT of the value.
Need the full syllabus wording and formula-booklet reference table? See Probability Distributions. For calculator steps, see the parent topic's GDC guidance.
Worked examples
The lifetime of a certain battery is modelled by \(X\sim N(500,40^2)\) hours.
Find the value of \(k\) such that \(P(500-k
Worked solution
By symmetry each tail has probability \(0.05,\) so \(P(X<500+k)=0.95.\) M1
\(z=\text{invNorm}(0.95)\approx1.6449.\) A1
\(k=1.6449\times40\) M1
\(\approx65.8.\) A1
\(X\sim N(40, \sigma^2)\) and \(P(X>50)=0.1.\) Find \(\sigma.\)
Worked solution
\(z\) for upper \(0.1\) is \(\approx 1.2816.\) M1 A1
\(1.2816 = \dfrac{50 - 40}{\sigma}\Rightarrow \sigma = \dfrac{10}{1.2816} \approx 7.80.\) A1
Exam scores are modelled by \(X\sim N(64,9^2).\) The top \(10\%\) of students receive a distinction, and the bottom \(20\%\) receive a fail.
(a) Find the minimum score needed for a distinction.
(b) Find the maximum score that results in a fail.
Worked solution
(a) Top 10% means \(P(X<k)=0.9,\ z=\text{invNorm}(0.9)\approx1.2816.\) M1 A1
\(k=64+1.2816(9)\approx75.5,\) so the minimum score for a distinction is \(76.\) A1
(b) \(z=\text{invNorm}(0.2)\approx-0.8416.\) M1 A1
\(k=64-0.8416(9)\approx56.4,\) so the maximum score for a fail is \(56.\) A1
Common mistakes
- Reporting the \(z\)-value instead of the actual \(x\)-value. Inverse normal calculations are asking for a real value in context (a mass, a time, a mark), not the intermediate standardized \(z\)-value used to find it.
- Entering the wrong tail probability. The GDC's inverse normal function always wants the area to the LEFT - if the question gives you an upper-tail or two-tail probability, convert it first or you'll get the wrong side of the distribution entirely.
- Rearranging the standardizing equation incorrectly. Once \(z\) is known, solving for \(\mu\) or \(\sigma\) means rearranging \(z=\dfrac{x-\mu}{\sigma}\) - a sign slip here (especially when \(z\) is negative) is one of the most common ways this topic loses marks.
Ready to practise properly?
11 inverse-normal questions, marked instantly like the real exam.
Quick answers
How do you find a value from a normal distribution when you're given a probability?
Use the GDC's inverse normal function with the area to the LEFT of the unknown value. It returns the \(x\)-value directly (or the \(z\)-value if you're working in standard form).
Why does inverse normal always need the area to the left?
That's how the function is defined on every GDC - it treats the number you enter as the cumulative probability up to that point. If you're given an upper-tail or two-tail probability, convert it to a left-tail probability before entering it.