Inverse normal

Find the value from a probability (inverse normal)

In 7 of 20 recent exams
Paper 1 or 2

How it is set in the exam

Paper 2 (calculator), usually 2 marks (up to 4 when the area has to be worked out first), straight after a normal probability part. Common (7 of 20 exam versions, 9 parts). Once on Paper 1 (no calculator), with no calculator: two normals with the same $\sigma$ .

Spot it in the exam

Exam phrases to spot (4)

“find the value of $w$ such that …”
“the heaviest $p\,\%$ … find the least mass”
“find $k$ such that $\mathrm{P}(X > \mu + k\sigma) = \ldots$ ”
“one tail is a multiple of the other”

In disguise (3)

"the top $10\%$ are awarded a distinction; find the lowest mark for a distinction": an upper tail, and its boundary
a part that only asks you to shade a region with an unknown bound, or to find $\mathrm{P}(X < w)$ , before the value itself is asked for: the set-up of this type
" $X$ and $Y$ have the same standard deviation", with a probability they share: equal distances from their means, on the same side for the same tail

Variants (3)

tail = left or upper; the accuracy demanded ("to the nearest gram", "to 4 significant figures")
given = the middle and one tail, a tail ratio, an interval probability and one of its bounds, or an equal probability in a second normal with the same $\sigma$ ()
on $Z$ : the answer is a number of standard deviations

Looks like it, but isn't (3)

$\mu$ or $\sigma$ is unknown and has to be found from probabilities at numerical values: Unknown mean or standard deviation
"the top $p\,\%$ " read off a cumulative frequency graph of data: Cumulative frequency graphs
a value is given and a probability asked for: Normal probabilities

Worked example

An upper tail, to a stated accuracy

Paper 2 (calculator)

Question

The masses of the parcels handled by a courier are normally distributed with mean $2.8$ kg and standard deviation $0.55$ kg. The heaviest $12\%$ of parcels are charged extra. Find the least mass of a parcel that is charged extra. Give your answer in kilograms, correct to two decimal places. [2]

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See it move: Find the boundary from the area

$w = 3.10$ : $\mathrm{P}(M > 3.10) = 0.293$ . More than 12% above: move right.
The heaviest 12% is an upper tail: 12% above $w$ , so 88% below it. That is the area inverse normal needs.
Inverse normal with area 0.88: $w = 3.4462\ldots \approx 3.45$ kg, above the mean as it must be.
Using 0.12 as the left area gives 2.15 kg, below the mean: impossible for the heaviest parcels.

Steps

Sketch
the curve and put the unknown on the correct side of $\mu$ : a left-tail area below $0.5$ puts it below $\mu$ . Shade the area you know.
2
Turn the information into one area to the left of the unknown:
- an upper tail $p$ : $1 - p$ ;
- a central region with probability $p$ , symmetric about $\mu$ : $\frac{1 - p}{2}$ to the left of its lower end, $\frac{1 + p}{2}$ to the left of its upper end;
- an interval whose lower end is the unknown: $1 - \text{interval} - \text{tail above it}$ ;
- an interval whose upper end is the unknown: $\text{tail below it} + \text{interval}$ (cumulate);
- a tail ratio: call the smaller tail $x$ , then $x + \text{middle} + kx = 1$ ; if the smaller tail is the lower one, the lower bound has $x$ to its left and the upper bound has $1 - kx$ .
Inverse normal
(GDC): write $\mathrm{P}(X < w) = \ldots$ , then $w$ . On with two normals that have the same $\sigma$ there is no GDC: equal areas on the same side put the two values the same distance from their own means (Standardisation).
Round as instructed
("to the nearest …", "to 4 s.f."), otherwise to 3 s.f. On $Z$ , the answer is the number of standard deviations itself.
Check
by putting the value back into the normal probability.

Common mistakes

An upper tail used as a left-tail area
for "the heaviest $12\%$ ", inverse normal on $0.12$ gives a value below the mean. The sketch shows it can't be right.
The unknown on the wrong side of $\mu$
in a sketch: the sketch mark needs the correct side and the area shown.
The wrong accuracy
"to the nearest gram", "correct to two decimal places" or "to 4 significant figures" is part of the final mark.

More mistakes (3)

Forgetting a tail
" $4x + 0.85 = 1$ " leaves out the left tail. All three regions go in the equation.
A $z$ -value as the answer
on $X$ the answer is a value in the question's units. Only when the question asks for a number of standard deviations (on $Z$ ) is $z$ itself the answer.
Two normals with the same $\sigma$
equal probabilities in the same tail mean equal distances from their own means, on the same side, not the same value.

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