Inverse Normal Calculations (College Board AP® Statistics)

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Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Inverse normal calculations

Given the value P(X < a) how do I find the value of a using the standard normal table?

  • The proportion of a distribution of a variable X that lies below a given value is the area under the curve to the left of the given value

    • This is also the probability that the variable lies below this value

    • straight P open parentheses X less than a close parentheses equals p

A normal distribution curve with mean μ and shaded area to the left of a vertical dashed line at a. The mean is labeled μ and point a is labeled a.
Normal distribution with proportion of distribution less than a shaded
  • straight P open parentheses X less than a close parentheses equals straight P open parentheses Z less than z close parentheses

    • where z is the z-score of the value a

  • If you are given the proportion (or probability) and want to work out the value of a:

    • find the cell in the standard normal table that is

      • equal to the proportion

      • or the highest value that is less than the proportion

    • identify the z-score by listing the relevant row and column

    • convert the z-score back into an actual value

Given the value of P(X < a) how do I find the value of a using a calculator?

  • Your calculator will have a function called Inverse Normal Distribution

    • Some calculators call this InvN

  • Given that straight P open parentheses X less than a close parentheses equals p you will need to enter:

    • The proportion (or probability), p

      • This is the area of the distribution to the left of a

    • The mean, mu

    • The standard deviation, sigma

  • Some calculators might ask for the tail

    • For straight P open parentheses X less than a close parentheses this is the left tail

Examiner Tips and Tricks

Always check your answer makes sense!

  • If straight P open parentheses X less than a close parentheses is less than 0.5 then a should be smaller than the mean,

  • If straight P open parentheses X less than a close parentheses is more than 0.5 then a should be bigger than the mean.

A sketch will help you see this.

Given the value of P(X > b) how do I find the value of a using the standard normal table?

  • The proportion of a a distribution of a variable X that lies above a given value is the area under the curve to the right of the given value

    • This is also the probability that the variable lies above this value

    • straight P open parentheses X greater than b close parentheses equals p

A normal distribution curve with mean μ and shaded area to the right of a vertical dashed line at b. The mean is labeled μ and point b is labeled b.
Normal distribution with proportion of distribution above b shaded
  • straight P open parentheses X greater than b close parentheses equals straight P open parentheses Z greater than z close parentheses

    • where z is the z-score of the value b

  • The total area under the curve is 1

    • So straight P open parentheses X greater than b close parentheses equals 1 minus straight P open parentheses X less than b close parentheses

  • If you are given the proportion (or probability) and want to work out the value of b:

    • subtract the proportion from 1

    • find the cell in the standard normal table that is

      • equal to this result

      • or the lowest value that is greater than the proportion

    • identify the z-score by listing the relevant row and column

    • convert the z-score back into an actual value

Given the value of P(X > b) how do I find the value of b using a calculator?

  • You will need to use the Inverse Normal Distribution function again

  • Given straight P open parentheses X greater than b close parentheses equals p

    • Use straight P open parentheses X less than b close parentheses equals 1 minus straight P open parentheses X greater than b close parentheses to rewrite this as straight P open parentheses X less than b close parentheses equals 1 minus p

    • Then use the method for straight P open parentheses X less than b close parentheses to find b

  • Your calculator may have the tail option (left, right or centre)

    • If so, you can use the Inverse Normal Distribution function straightaway by:

      • selecting right for the tail

      • and entering the area as p

Examiner Tips and Tricks

Always check your answer makes sense!

  • If straight P open parentheses X greater than b close parentheses is less than 0.5 then a should be bigger than the mean,

  • If straight P open parentheses X greater than b close parentheses is more than 0.5 then a should be smaller than the mean.

A sketch will help you see this.

Worked Example

The distribution of heights of all adult male bison in a particular herd of bison is approximately normal with a mean of 5.9 feet and standard deviation 0.16 feet. The rancher wants to select the tallest twenty percent of the bison to use for a breeding program.

What is the minimum height that an adult male bison must be to qualify for the breeding program?

Answer:

Method 1: Using the standard normal table

Draw a sketch of the situation

Because P open parentheses X greater than b close parentheses equals 0.2, is less than 0.5, b should be bigger than the mean, i.e. b lies to the right of the mean

A normal distribution with the mean, μ, at 6.8. The area under the curve to the right of 'b' is highlighted and has probability, P(X > b) = 0.2.

Using the standard normal table, look at section with the positive z-values as b is above the mean

The standard normal table contains the proportions or percentages that are below a given value, i.e. to the left of the given value

You are given the proportion that lies to the right of b, so use P open parentheses X less than b close parentheses equals 1 minus P open parentheses X greater than b close parentheses to find the proportion to the left of b

1 minus 0.2 equals 0.8

Find the cell that contains the value that is 0.8, if there is no exact value, find the smallest value that is greater than 0.8

0.8023 is the lowest value in the standard normal table that is greater than 0.8

Write down the z-score from the corresponding row and column

z equals 0.85

Convert the z-score into an actual value, using z equals fraction numerator x minus mu over denominator sigma end fraction

table row cell 0.85 end cell equals cell fraction numerator b minus 5.9 over denominator 0.16 end fraction end cell row cell 0.85 times 0.16 end cell equals cell b minus 5.9 end cell row cell 0.85 times 0.16 plus 5.9 end cell equals b row b equals cell 6.036 end cell end table

Explain the value in the context of the question

The minimum height that an adult male bison must be to qualify for the breeding program is 6.04 feet

Method 2: Using a calculator

Draw a sketch of the situation

Because P open parentheses X greater than b close parentheses equals 0.2, it is less than 0.5 , therefore b should be bigger than the mean, i.e. to the right of the mean

A normal distribution with the mean, μ, at 6.8. The area under the curve to the right of 'b' is highlighted and has probability, P(X > b) = 0.2.

You are given the proportion that lies to the right of b, so use P open parentheses X less than b close parentheses equals 1 minus P open parentheses X greater than b close parentheses to find the proportion to the left of b

1 minus 0.2 equals 0.8

Write down the parameters for the situation

straight P open parentheses X less than b close parentheses equals 0.8

mu equals 5.9

sigma equals 0.16

Enter these values into the Inverse Normal Distribution function on your calculator and calculate b

b equals 6.036

Explain the value in the context of the question

The minimum height that an adult male bison must be to qualify for the breeding program is 6.04 feet

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Naomi C

Author: Naomi C

Expertise: Maths

Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.