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 $X\sim N(\mu ,{\sigma }^2)$ and a probability $p$, the inverse normal distribution function finds a value $k$​ such that $P(X\le k)=p$​.

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.

​$$A = P(X \le k)$$​​

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Using the inverse normal distribution

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

calculator tip

$\boxed{\begin{array}{rl}& Menu\\ 7& :Distribution\\ 3& :InverseNormal\\ Area& :\\ \sigma & :\\ \mu & :\end{array}}$​​

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Example 1

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

i) $a$ such that $P(X\le a)=0.7$.

ii) $b$ such that $P(X\ge b)=0.23$.

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

Part i):

​ ​	​$\boxed{\begin{array}{rl}& Menu\\ 7& :Distribution\\ 3& :InverseNormal\\ Area& :0.7\\ \sigma & :0.5\\ \mu & :1\end{array}}$​​
	$$\begin{gathered} \boxed{xInv= \\ 1.2622} \end{gathered}$$​​

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

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:

​ ​ ​ ​	​$$\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.77\\ \sigma&:0.5\\\mu &:1 \end{aligned}}$$​​
​	​$$\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.3694}$$​​

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

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:

​ ​ ​	​$$\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.6319\\ \sigma&:0.5\\\mu &:1 \end{aligned}}$$​​
	​$$\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.1684}$$​​

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