Inverse Normal Calculations (College Board AP® Statistics): Study Guide

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

  • $\mathrm{P}\left(X<a\right)=p$

• $\mathrm{P}\left(X<a\right)=\mathrm{P}\left(Z<z\right)$

  • 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 $\mathrm{P}\left(X<a\right)=p$ 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 $\mathrm{P}\left(X<a\right)$ this is the left tail

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

  • $\mathrm{P}\left(X>b\right)=p$

• $\mathrm{P}\left(X>b\right)=\mathrm{P}\left(Z>z\right)$

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

• The total area under the curve is 1

  • So $\mathrm{P}\left(X>b\right)=1-\mathrm{P}\left(X<b\right)$

• 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 $\mathrm{P}\left(X>b\right)=p$

  • Use $\mathrm{P}\left(X<b\right)=1-\mathrm{P}\left(X>b\right)$ to rewrite this as $\mathrm{P}\left(X<b\right)=1-p$

  • Then use the method for $\mathrm{P}\left(X<b\right)$ 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 $p$

How can I use the table for t-distribution critical values?

• You will later learn that the t-distribution tends to the standard normal distribution when the degree of freedom tends to infinity

• In your exam, you have a table of t-distribution critical values

  • These give the values of $t$ such that $P\left(T>t\right)=p$

• You can use the row for infinity (∞) to find z-scores that give probabilities $P\left(Z>z\right)=p$

• The table only gives certain common values of $p$

• You can use symmetries of the normal distribution to find negative z-scores

  • For example, the table tells you $P\left(Z>z\right)=0.025\Rightarrow z=1.960$

    • This means $P\left(Z<z\right)=0.025\Rightarrow z=-1.960$

• To use this table, write the probability statement in the form $P\left(Z>z\right)=p$

  • If the probability is greater than 0.5, subtract it from 1 and flip the inequality

    • For example, if you want $P\left(Z<z\right)=0.9$ then rewrite as $P\left(Z>z\right)=0.1$

  • If the sign is $<$, then flip the inequality and change the sign of the z-score

    • For example, if you want $P\left(Z<z\right)=0.1$ then find $P\left(Z>z\right)=0.1$ and change the z-score to a negative