The inverse normal distribution function

In a nutshell

The inverse normal distribution function finds a value that corresponds to a known probability. In other words, given $\sim N(\mu, \sigma^2)$ and a probability $p$ , the inverse normal distribution function finds a value $k$ such that $\leq k )=p$ .

Area and probability

Probabilities for a normal distribution are equal to the corresponding area under the graph of the corresponding distribution. In particular, the cumulative probability $P(X \le k)$ is equal to the area of the region that is to the left of the line $x=k$ , under the graph.

Maths; Normal distribution; KS5 Year 13; The inverse normal distribution function

A = P(X \le k)

Using the inverse normal distribution

The inverse normal distribution function can be used on your calculator.

Maths; Normal distribution; KS5 Year 13; The inverse normal distribution function

calculator tip

$\boxed{\begin{aligned} & \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&: Inverse \ Normal \\Area&:\\\sigma&:\\ \mu&: \end{aligned}}$

Example 1

Given the random variable $\sim N(1,0.25)$ , find the following values to two decimal places.

i) $a$ such that $\leq a) = 0.7$ .

ii) $b$ such that $\geq b ) = 0.23$ .

iii) $c$ such that $P(c \leq X \lt 1.04)=0.1$ .

Part i):

Maths; Normal distribution; KS5 Year 13; The inverse normal distribution function

$\boxed{\begin{aligned} & \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&: Inverse \ Normal \\Area&:0.7\\\sigma&:0.5\\ \mu&:1 \end{aligned}}$

$\boxed{xInv = \\ \ \\ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ 1.2622 }$

$a = \underline{1.26 \ (2 \ d.p.)}$

Part ii):

Rearrange this into the form $P(X \leq ...)=...$ :

$\begin{aligned}P(X \geq b) &= 0.23\\1 - P(X \lt b)&= 0.23\\P(X\lt b) = P(X \le b)&= 0.77 \end{aligned}$ 

Now, use the inverse normal distribution:

Maths; Normal distribution; KS5 Year 13; The inverse normal distribution function

$\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.77\\ \sigma&:0.5\\\mu &:1 \end{aligned}}$

$\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.3694}$

$b = \underline{1.40 \ (2 \ d.p.)}$

Part iii):

Rearrange:

$\begin{aligned} P(c \leq X \lt 1.04)&=0.1\\P(X \lt 1.04) - P(X \leq c)&=0.1\\P(X \le c) &=0.1+P(X \lt 1.04)\\P(X \leq c) &=0.1+0.5319\\P(X\le c)&=0.6319\end{aligned}$ 

Use the inverse normal distribution:

Maths; Normal distribution; KS5 Year 13; The inverse normal distribution function

$\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.6319\\ \sigma&:0.5\\\mu &:1 \end{aligned}}$

\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.1684}

$c= \underline{1.17 \ (2 \ d.p.)}$