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Normal distribution

The inverse normal distribution function

The inverse normal distribution function

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Explainer Video

Tutor: Bilal

Summary

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 ∼ N ( μ , σ 2 ) X \sim N(\mu, \sigma^2) X∼N(μ,σ2) and a probability p p p, the inverse normal distribution function finds a value k k k​ such that P ( X ≤ k ) = p P(X \leq k )=p P(X≤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≤k)P(X \le k)P(X≤k) is equal to the area of the region that is to the left of the line x=kx=kx=k, under the graph.


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

​A=P(X≤k)A = P(X \le k)A=P(X≤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

        M e n u 7 : D i s t r i b u t i o n 3 : I n v e r s e   N o r m a l A r e a : σ : μ : \boxed{\begin{aligned} & \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&: Inverse \ Normal \\Area&:\\\sigma&:\\ \mu&: \end{aligned}} 73Areaσμ​       Menu:Distribution:Inverse Normal:::​​​​

​​

Example 1

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

i) a a a such that P ( X ≤ a ) = 0.7 P(X \leq a) = 0.7 P(X≤a)=0.7.

ii) b b b such that P ( X ≥ b ) = 0.23 P(X \geq b ) = 0.23 P(X≥b)=0.23.

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


Part i):

​



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


​

​         M e n u 7 : D i s t r i b u t i o n 3 : I n v e r s e   N o r m a l A r e a : 0.7 σ : 0.5 μ : 1 \boxed{\begin{aligned} & \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&: Inverse \ Normal \\Area&:0.7\\\sigma&:0.5\\ \mu&:1 \end{aligned}} 73Areaσμ​       Menu:Distribution:Inverse Normal:0.7:0.5:1​​​​

x I n v =                    1.2622 \boxed{xInv = \\ \ \\ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ 1.2622 } xInv=                1.2622​​​


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


Part ii):

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

P(X≥b)=0.231−P(X<b)=0.23P(X<b)=P(X≤b)=0.77\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}P(X≥b)1−P(X<b)P(X<b)=P(X≤b)​=0.23=0.23=0.77​​​


Now, use the inverse normal distribution:

​

​


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


​

​       Menu7:Distribution3:Inverse NormalArea:0.77σ:0.5μ:1\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.77\\ \sigma&:0.5\\\mu &:1 \end{aligned}}73Areaσμ​       Menu:Distribution:Inverse Normal:0.77:0.5:1​​​​

​

​xInv=                1.3694\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.3694}xInv=                1.3694​​​


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


Part iii):

Rearrange:

P(c≤X<1.04)=0.1P(X<1.04)−P(X≤c)=0.1P(X≤c)=0.1+P(X<1.04)P(X≤c)=0.1+0.5319P(X≤c)=0.6319\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}P(c≤X<1.04)P(X<1.04)−P(X≤c)P(X≤c)P(X≤c)P(X≤c)​=0.1=0.1=0.1+P(X<1.04)=0.1+0.5319=0.6319​​​


Use the inverse normal distribution:

​


​


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


​

​       Menu7:Distribution3:Inverse NormalArea:0.6319σ:0.5μ:1\boxed{\begin{aligned}& \ \ \ \ \ \ \ Menu\\7&: Distribution\\3&:Inverse \ Normal\\Area&:0.6319\\ \sigma&:0.5\\\mu &:1 \end{aligned}}73Areaσμ​       Menu:Distribution:Inverse Normal:0.6319:0.5:1​​​​
​xInv=                1.1684\boxed{xInv = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.1684}xInv=                1.1684​​​


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




 

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The normal distribution

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The normal distribution

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FAQs - Frequently Asked Questions

What type of cumulative probabilities does the inverse normal distribution function work for?

The inverse normal distribution function works for probabilities of the form P(X < a) or P(X ≤ a).

What is the inverse normal distribution function?

The inverse normal distribution finds values that correspond to a given cumulative probability.

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