13 3invrsnml

Inverse Cumulative Normal Distribution

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Aim: To answer questions of the form: Find a so that $Pr\big(X < a\big) = 0.9$
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Historical Note:

• Before calculators, to find a we would have used the CND table backwards.
• ie Look in the body of the table to find the closest entry to 0.9 and then read off the z-value
• From the table: $Pr (Z < b) = 0.9$ gives $b = 1.28$ (then convert from Z to X)

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The Good News

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Use of the CND table is no longer in the course. Now we get to use calculators.
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… … These problems can be solved using the Inverse Cumulative Normal Distribution.

… … Like all Inverse functions, it allows us to go backwards

… …. … in the same way that $\sin^{-1}(x)$ is the inverse of $\sin(x)$
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… … You will find the InvNormCDf function in the Distribution/Inverse entry of the Interactive menu.

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Tail

… … The first option on the menu for InvNormCDf is "Tail"

… … Using $X \sim N\big(\mu, \; \sigma^2\big)$
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… … $\text{InvNormCDf}\big(Left,\; p, \; \sigma, \; \mu\big)$

… … supplies a where $Pr\big(X < a\big) = p$

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… … $\text{InvNormCDf}\big(Right,\; p, \; \sigma, \; \mu\big)$

… … supplies a where $Pr\big(X > a\big) = p$

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… … $\text{InvNormCDf}\big(Center,\; p, \; \sigma, \; \mu\big)$

… … supplies a where $Pr\big(a < X < b \big) = p$

… … {Where a and b are symmetrical about $\mu$ }

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Note: Some types of calculators can only calculate the Left Tail.

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Example 1

a) .. Find a so that $Pr\big(Z < a\big) = 0.9$

b) .. Given $X \sim N\big(10, \; 9\big)$, find b so that $Pr\big(X < b\big) = 0.7$

c) .. Given $X \sim N\big(10, \; 9\big)$, find c so that $Pr\big(X > c\big) = 0.65$

d) .. Given $X \sim N\big(12, \; 16\big)$, find a so that $Pr\big(12 - a < X < 12 + a\big) = 0.4$

*Solution**

a) .. Find a so that $Pr\big(Z < a\big) = 0.9$
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… … … Recall that $Z \sim N\big(0, \; 1\big)$
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… … … enter: $\text{InvNormCDf, } Tail = Left, \; p = 0.9, \; \sigma = 1, \; \mu = 0$
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… … … $a = 1.2816$

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b) .. Given $X \sim N\big(10, \; 9\big)$, find b so that $Pr\big(X < b\big) = 0.7$
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… … … enter: $\text{InvNormCDf, } Tail = Left, \; p = 0.7, \; \sigma = 3, \; \mu = 10$
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… … … $b = 11.5732$

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c) .. Given $X \sim N\big(10, \; 9\big)$, find c so that $Pr\big(X > c\big) = 0.65$
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… … … enter: $\text{InvNormCDf, } Tail = Right, \; p = 0.65, \; \sigma = 3, \; \mu = 10$
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… … … $c = 8.8440$

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Note: If your calculator only finds the Left tail:
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… … … $Pr\big(X > c\big) = 0.65$
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… … … $Pr\big(X < c\big) = 1 -0.65$
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… … … $Pr\big(X < c\big) = 0.35$
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… … … enter: $\text{InvNormCDf, } Tail = Left, \; p = 0.35, \; \sigma = 3, \; \mu = 10$
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… … … $c = 8.8440$

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d) .. Given $X \sim N\big(12, \; 16\big)$, find a so that $Pr\big( 12-a < X < 12+a \big) = 0.4$
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… … … enter: $\text{InvNormCDf, } Tail = Center, \; p = 0.4, \; \sigma = 4, \; \mu = 12$
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… … … Note: This returns $9.9024$ which represents the left or lower limit of our desired area.
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… … … $12 - a = 9.9024$
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… … … $a = 12 - 9.9024$
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… … … $a = 2.0976$

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… … … The right or upper limit is therefore $12 + 2.0976 = 14.0976$
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Note: If your calculator only finds the Left tail:
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… … … $Pr \big( 12-a < X < 12+a \big) = 0.4$
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… … … $Pr\Bigg( \big( X < 12-a \big) \cup \big( X > 12+a \big) \Bigg) = 1 - 0.4 = 0.6$
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… … … $Pr \big( X < 12 - a \big) = \dfrac{0.6}{2} = 0.3$

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… … … The calc will give the left limit value of $9.9024$
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… … … $12 - a = 9.9024$
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… … … $a = 2.0976$

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Finding an Unknown Mean or Standard Deviation

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This can be done by finding the appropriate Z value that gives the desired probability

Then using the rule to convert from Z back to X

… … $z=\dfrac{x - \mu}{\sigma}$

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Example 2

Find $\sigma$, given that $X \sim N\big(9, \; \sigma^2\big)$, and $Pr\big(X < 12\big) = 0.8$,

Solution:
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… … {We don't know the standard deviation so we can't use the calculator immediately}
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… … We do know that $Z \sim N\big(0, \; 1\big)$
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… … so we can find a so that $Pr\big(Z < a\big) = 0.8$
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… … … enter: $\text{InvNormCDf, } Tail = Left, \; p = 0.8, \; \sigma = 1, \; \mu = 0$
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… … This returns $a = 0.8416$
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… … Hence $Pr\big(X < 12\big) = Pr\big(Z < 0.8416\big) = 0.8$
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… … So when $x = 12$ we have that $z = 0.8416$,
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… … Now we can find $\sigma$ using the equation:
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… … $z=\dfrac{x - \mu}{\sigma}$
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… … $0.8416 = \dfrac{12 - 9}{\sigma}$
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… … $\sigma = \dfrac{12-9}{0.8416}$
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… … $\sigma = 3.56$

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