13 2stndrdnml

# Standard Normal Distribution

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Recall that a Normal Distribution is denoted by: $X \sim N\big(\mu,\; \sigma^2\big)$
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## Historical Note

• Before calculators, to find a probability other than the confidence intervals involved integrating the normal pdf curve for those particular values of mean and standard deviation. This was long and time consuming. Thus the Standard Normal Distribution was created.

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• The Standard Normal Distribution has a mean of 0 and a standard deviation of 1

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• Any normal distribution can be converted into the standard normal distribution.

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• Then, instead of integrating, a table of values for the standard normal distribution would be consulted.

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• This table is called the CND table (Cumulative Normal Distribution)

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## Standard Normal Distribution

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A standard normal Distribution has $\mu = 0 \text{ and } \sigma = 1$.
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We use the variable Z (instead of X) to indicate a variable in the Standard Normal Distribution.
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Therefore: … $Z \sim N\big(0,\; 1\big)$

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Any value, x, in a normal distribution $X \sim N\big(\mu,\; \sigma^2\big)$, can be converted into the equivalent z value in a standard normal distribution, using:

… … $z=\dfrac{x-\mu}{\sigma}$

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Historically, $Pr(Z < z)$ would then be looked up on a CND table.
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Now we can simply use technology to find either $Pr\big(X < x\big)$ or $Pr\big(Z < z\big)$
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### Example 1

Given $X \sim N\big(84,\; 16\big)$ ,

Convert $X = 86$ to the standard normal distribution, Z

hence find $Pr \big(X < 86\big)$

Solution

… … Note: $\sigma^2 = 16 \; \text{ so } \; \sigma = 4$
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… … $z = \dfrac{x -\mu}{\sigma}$
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… … $z = \dfrac{86-84}{4}$
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… … $z = \dfrac{2}{4}$
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… … $z = 0.5$
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… … so
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… … $Pr(X < 86)$
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… … … $= Pr(Z < 0.5)$
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… … … $= 0.6915$ … {from CND table or technology}
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## GOOD NEWS

Using the CND table is no longer part of the course.
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You are still expected to be able to convert between a value x in a normal distribution and the equivalent value z in the standard normal distribution using the rule:

… … $z=\dfrac{x-\mu}{\sigma}$
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You are expected to use your calculator to find probabilities for both a Normal Distribution X and a Standard Normal Distribution Z.

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### Example 1 (continued)

Given $X \sim N\big(84,\; 16\big)$

Use CAS to find $Pr\big(X < 86\big)$

Solution

… … $Pr\big(X < 86\big)$
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… … … enter: $\text{normalcdf Lr } = -\infty, \; \text{ Ur } = 86, \; \sigma = 4, \; \mu = 84$
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… … … $Pr\big(X < 86\big) = 0.6915$
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… … OR
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… … $z = \dfrac{86-84}{4}$
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… … $z = 0.5$
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… … so
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… … $Pr\big(X < 86\big)$
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… … … $= Pr\big(Z < 0.5\big)$
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… … … enter: $\text{normalcdf Lr } = -\infty, \; \text{ Ur } = 0.5, \; \sigma = 1, \; \mu = 0$
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… … … $Pr\big(Z < 0.5\big) = 0.6915$
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… … so
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… … $Pr\big(X < 86\big) = 0.6915$

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Note: Don't write calculator notation as a part of your answer in tests/exams.

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### Example 2

Use CAS to find
a) .. $Pr\big(Z > 0.4\big)$

b) .. $Pr\big(0.2 < Z < 1.1\big)$

Solution

a) .. $\text{Find } Pr\big(Z > 0.4\big)$
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… … … enter: $\text{normalcdf Lr } = 0.4, \; \text{ Ur } = \infty, \; \sigma = 1, \; \mu = 0$
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… … … $Pr \big(Z > 0.4\big) = 0.3446$

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b) .. $\text{Find } Pr\big(0.2 < Z < 1.1\big)$
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… … … enter: $\text{normalcdf Lr } = 0.2, \; \text{ Ur } = 1.1, \; \sigma = 1, \; \mu = 0$
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… … … $Pr\big(0.2 < Z < 1.1\big) = 0.2851$

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NOTE I recommend using the Interactive version so that you are prompted for the values in the correct order.
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NOTE Do NOT use normalpdf at all (ever)

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## Properties of Normal Distributions

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We already know that for any distribution (including Normal Distributions), it is true that

… … $Pr\big(X > a\big) = 1 - Pr\big(X < a\big)$
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All Normal Distributions (including the Standard Normal Distribution) are symmetrical
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This means that:

… … $Pr\big(X > a\big) = Pr\big(X < -a\big)$ … {See diagram}

… … or

… … $Pr\big(X < a\big) = Pr\big(X > –a\big)$
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Consideration of the second diagram shows that $\big( \text{if } a < b \big)$

… … $Pr\big(a < X < b\big) = Pr\big(X < b\big) - Pr\big(X < a\big)$

… … and

… … $Pr \Big( \, (X < a) \cap Pr(X < b) \, \Big) = Pr\big(X < a\big)$

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### Example 3

{Calculator Free Question}

Given
… … $Pr\big(Z < a\big) = 0.2$

… … $Pr\big(Z < b\big) = 0.6$

Find

a) .. $Pr\big(Z > a\big)$

b) .. $Pr\big(Z < -b\big)$

c) .. $Pr\big(a < Z < b\big)$

d) .. $Pr\big(Z < a \,|\, Z < b\big)$

Solution

a) .. $\text{Find } Pr\big(Z > a\big)$
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… … … $Pr(Z > a)$
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… … … … $= 1 – Pr(Z < a)$
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… … … … $= 1 – 0.2$
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… … … … $= 0.8$
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b) .. $\text{Find } Pr\big(Z < -b\big)$
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… … … $Pr(Z < -b)$
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… … … … $= Pr(Z > b)$
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… … … … $= 1 - Pr(Z < b)$
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… … … … $= 1 – 0.6$
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… … … … $= 0.4$

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c) .. $\text{Find } Pr\big(a < Z < b\big)$
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… … … $Pr(a < Z < b)$
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… … … … $= Pr(Z < b) - Pr(Z < a)$
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… … … … $= 0.6 - 0.2$
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… … … … $= 0.4$

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d) .. $\text{Find } Pr\big(Z < a \,|\, Z < b\big)$
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… … … $Pr \big( Z < a \, | \, Z < b \big)$
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… … … … $= \dfrac{Pr \big( (Z < a) \cap (Z < b) \big) }{Pr(Z < b)}$
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… … … … $= \dfrac{Pr(Z < a)}{Pr(Z < b)}$
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… … … … $= \dfrac{0.2}{0.6}$
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… … … … $= \dfrac{1}{3}$

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### Don't Forget!

… … $Pr(Z < a) = Pr(Z \leqslant a)$

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