Mathematics/Lessons/The Normal Distribution/Inverse Normal Distribution Function
Statistics · The Normal Distribution · Year 2

Inverse Normal Distribution Function

Working backwards from a known area to the value of x that produces it.

8 min read 2 worked examples

Working backwards from a probability

For a given probability $p$, you can use your calculator to find a value $a$ such that $\mathrm{P}(X<a)=p$. This is usually called the inverse normal distribution function on your calculator.

Worked example 1

Finding a from a given area
$X\sim \mathrm{N}(52, 5^2)$. Find, correct to two decimal places, the values of $a$ such that: a $\mathrm{P}(X<a)=0.65$   b $\mathrm{P}(X>a)=0.3$   c $\mathrm{P}(45<X<a)=0.4$

a

37 42 47 52 53.93 57 62 67 X ~ N(52, 5²)
$a=53.93$ (2 d.p.). $0.65$ is more than $0.5$, so $a$ should be greater than the mean.

Enter $\mu=52$, $\sigma=5$ and $p=0.65$ into your calculator. This means that for $X\sim \mathrm{N}(52,5^2)$, $\mathrm{P}(X<53.93)=0.65$.

$a = 53.93$ (2 d.p.).

b

37 42 47 52 54.62 57 62 67 X ~ N(52, 5²)
$a=54.62$. $\mathrm{P}(X>a)=0.3$ means $\mathrm{P}(X<a)=0.7$.
1
$\mathrm{P}(X>a)=0.3$, so $\mathrm{P}(X<a)=1-0.3=0.7$.
Use the fact that $\mathrm{P}(X>a)+\mathrm{P}(X<a)=1$ to find the area to the left of $a$ before using your calculator.
2
$a = 54.62$ (2 d.p.).
$\mathrm{P}(X<54.62)=0.7$, so $\mathrm{P}(X>54.62)=0.3$ as needed.
▪ solved

c

37 42 45 51.76 52 57 62 67 X ~ N(52, 5²)
$a=51.76$. The unknown value $a$ is above the mean, since the interval already accounts for $0.4$ of the probability starting from below the mean.
MethodConvert to a "less than" statement first
1
$\mathrm{P}(45<X<a) = \mathrm{P}(X<a)-\mathrm{P}(X<45)$.
You can't enter $\mathrm{P}(45<X<a)=0.4$ directly — the inverse function only works on $\mathrm{P}(X<a)=p$.
2
$\mathrm{P}(X<45) = 0.0808$, so $\mathrm{P}(X<a) = 0.4+0.0808 = 0.4808$.
Find $\mathrm{P}(X<45)$ first using the normal cumulative distribution function.
3
$a = 51.76$ (2 d.p.).
You can check this using $\mathrm{P}(45<X<51.76)$.
▪ solved

Worked example 2

Percentiles and the interquartile range
The length, $L$ mm, of screws produced by a machine is modelled as $L\sim \mathrm{N}(45, 2^2)$. a Given that 55% of screws are shorter than $x$ mm, find $x$. b Find the interquartile range of the screw lengths.

a

40 42 45 45.25 48 50 L ~ N(45, 2²)
$\mathrm{P}(L<x)=0.55 \Rightarrow x=45.25$mm.

$\mathrm{P}(L<x)=0.55 \Rightarrow x = 45.25$ mm.

b

40 42 43.65 45 46.35 48 50 L ~ N(45, 2²)
The lower quartile $Q_1=43.65$mm and upper quartile $Q_3=46.35$mm are equidistant from the mean, since the distribution is symmetrical.
1
$\mathrm{P}(L<Q_1)=0.25 \Rightarrow Q_1 = 43.65$ mm.
25% of the data values lie below the lower quartile.
2
$\mathrm{P}(L<Q_3)=0.75 \Rightarrow Q_3 = 46.35$ mm.
75% of the data values lie below the upper quartile.
3
The interquartile range is $46.35-43.65 = 2.70$ mm (2 d.p.).
The distribution is symmetrical, so $Q_1$ and $Q_3$ should be the same distance from the mean — here, both are $1.35$mm from $\mu=45$.
▪ solved

Check yourself

The time, $T$ minutes, taken by students to complete a puzzle is modelled as $T\sim \mathrm{N}(25, 4^2)$. Given that 20% of attempts take longer than $t$ minutes, find $t$.

1
$\mathrm{P}(T>t)=0.2 \Rightarrow \mathrm{P}(T<t)=0.8$.
2
$t = 28.37$ minutes (2 d.p.).
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