11

Normal Random VariablesNormal Random Variables

In the class of continuous random variables, we are In the class of continuous random variables, we are primarily interested in NORMAL random variables. primarily interested in NORMAL random variables.

These are a continuous random variable with a bell-These are a continuous random variable with a bell-shaped distribution. These normal or bell-shaped shaped distribution. These normal or bell-shaped variables occur often in nature.variables occur often in nature.

Example:Example: Heights of MalesHeights of Males

are Normally Distributedare Normally Distributed

Probability Density FunctionProbability Density Function

forfor Heights of Males --->Heights of Males ---> 55 60 65 70 75 80

Height (inches)

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Properties of the Normal Properties of the Normal distributiondistribution

There are infinitely many normal pdf’s (curves). To There are infinitely many normal pdf’s (curves). To fully describe a normal curve, we need the location fully describe a normal curve, we need the location (mean, (mean, μμ) and the spread (standard deviation, ) and the spread (standard deviation, σσ). ).

NotationNotation: : If X a normal random variable with mean EIf X a normal random variable with mean E(X)=(X)=μ μ and variance Varand variance Var(X)=(X)=σσ22 we writewe write

XX~N (~N (μμ, , σσ22)) For population mean and s.d. we use the Greek For population mean and s.d. we use the Greek letters letters μμ and and σσ, for the sample mean and s.d. we , for the sample mean and s.d. we use and use and ss..

The distribution is symmetrical around the mean The distribution is symmetrical around the mean μμ.. The median, and the mean are equal due to the The median, and the mean are equal due to the

symmetry of the distribution.symmetry of the distribution. The total area under the curve is equal to one.The total area under the curve is equal to one.

x

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The parameters The parameters µµ and and σσ

µ=-2 µ=2µ=0

For all, σ=1 For all, µ =0

σ=2

σ=1/2

σ=1

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Probabilities with Normal RVsProbabilities with Normal RVs

When we consider Normal Random Variables (or When we consider Normal Random Variables (or any continuous r.v.), we are interested in the any continuous r.v.), we are interested in the probability that Xprobability that X falls into some INTERVAL.falls into some INTERVAL.

The probability a random variable The probability a random variable XX~N~N((μμ, , σσ22)) to to take a value is equal to zero. In other words, if take a value is equal to zero. In other words, if X~NX~N((μμ, , σσ22)) then then P(X=k)=0P(X=k)=0, where , where kk is some is some number.number.

Example:Example: Suppose X is the height of a randomly Suppose X is the height of a randomly chosen college woman. Further suppose that the chosen college woman. Further suppose that the heights of college women can be described as a heights of college women can be described as a normal, with normal, with μμ = 65inches (in), and = 65inches (in), and σσ = 2.7 in. = 2.7 in.

We might ask:We might ask:1.1. What is the proportion of women that are shorter What is the proportion of women that are shorter

than 62 in?than 62 in?2.2. What is the probability that What is the probability that XX is between 65 and is between 65 and

67in?67in?

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Graphical Representation of Graphical Representation of ProbabilitiesProbabilities

P(X<62)

P(65<X<67)

The total area under the curve is equal to 1! The probability that X falls in an interval is equal to the

area of the region below the curve and over the interval.

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Probabilities with Normal RVsProbabilities with Normal RVs

The total area under the curve is equal to 1! The probability that X falls in an interval is equal

to the area of the region below the curve and over the interval.

For example For example P(a P(a X X b) b) is equal to is equal to the area the area under the curve between under the curve between aa and and bb..

Due to the continuity of the normal distribution Due to the continuity of the normal distribution we have that the probability a normal random we have that the probability a normal random variable to take a value is equal to zero, thus variable to take a value is equal to zero, thus

P(X P(X a)= P(X a)= P(X < < a) a) P(a P(a X X b)= P(a < X b)= P(a < X < < b) b) P(a P(a X X << b)= P(a < X b)= P(a < X < < b)b) etc. etc.

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Standard Normal DistributionStandard Normal Distribution

We have tables for the probabilities of the form We have tables for the probabilities of the form P(Z < z) P(Z < z) where where z z ≥≥ 00. .

e.g. e.g. P(Z ≤ 0.5), P(Z < 2P(Z ≤ 0.5), P(Z < 2). ). Probabilities of the form Probabilities of the form P(Z < - 0.4), P(Z > 1.2), P(Z < - 0.4), P(Z > 1.2),

P(Z > - 0.25) P(Z > - 0.25) have to be transformed into have to be transformed into probabilities of the form probabilities of the form P(Z < z) P(Z < z) where where z z ≥≥ 0. 0.

A normal r.v. with A normal r.v. with µµ=0 =0 and and σσ=1 is called a standard =1 is called a standard normal random variable.normal random variable. We denote it with We denote it with Z, Z, so so that that

ZZ~N(0,1).~N(0,1).

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How to use the How to use the tabletable

The probability The probability P(Z<1.14)P(Z<1.14) is the is the number in the table number in the table where the row of where the row of 1.1 and the column 1.1 and the column of .04 are crossed. of .04 are crossed. Thus, Thus, P(Z<1.14)=0.8729P(Z<1.14)=0.8729

More examples: More examples: P(Z<0.57)=.7157 P(Z<0.57)=.7157 P(Z<2)=.9972 P(Z<2)=.9972 P(Z<1.3)=.9032 P(Z<1.3)=.9032 P(Z<0)=0.5P(Z<0)=0.5

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Calculating Probabilities of Calculating Probabilities of ZZ~N(0,1)~N(0,1)

For For ZZ~N(0,1), and ~N(0,1), and αα ≥≥0:0:

1.1. P(Z > P(Z > αα) =1-P() =1-P(Z < Z < αα))

2.2. P(Z < -P(Z < -αα) =1-P() =1-P(Z < Z < αα))

3.3. P(Z > -P(Z > -αα) =P() =P(Z < Z < αα))

4.4. P(b <Z < cP(b <Z < c) =P() =P(Z < Z < c) - P(c) - P(Z < Z < b),b), for any for any bb and and c.c.

Draw a normal curve and shade the areas corresponding Draw a normal curve and shade the areas corresponding to the above probabilities.to the above probabilities.

1010

. of score-z the called is , where

that have weresult this Using

xx

z

z),P(Zσ

μxσ

μXPx)P(X

Calculating Probabilities of any Calculating Probabilities of any normal r.v. X~N( normal r.v. X~N( µ , µ , σσ ))

).,(~ i.e.

on,distributi normal standard a has then ),,(~ If

10

2

NX

Z

XNX

We can obtain any type of probabilities of interest for We can obtain any type of probabilities of interest for any normal r.v any normal r.v XX~N(~N(μ, σ μ, σ 22) by first transforming ) by first transforming XX into into Z Z using the following using the following “standardization theorem’“standardization theorem’::

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How to Calculate ProbabilitiesHow to Calculate Probabilities

If you want P(X < x), first compute the z-score:If you want P(X < x), first compute the z-score:

z = (Value – mean)/(Standard Deviation) = (x-z = (Value – mean)/(Standard Deviation) = (x-µ)/µ)/σσ, ,

But P(X < x) = P(Z < z) for which we have tables!!But P(X < x) = P(Z < z) for which we have tables!!

ExampleExample: : X X = height of a college woman, = height of a college woman, XX~N( 65, ~N( 65, 2.72.722))

1.1. P(X < 62)P(X < 62) z = (62 – 65)/2.7 = -3/2.7 = -1.11z = (62 – 65)/2.7 = -3/2.7 = -1.11 P(X < 62) = P(Z < -1.11) (now use Normal Table) P(X < 62) = P(Z < -1.11) (now use Normal Table) = 0.134 = 13.4%= 0.134 = 13.4%

2.2. P(65 < X < 67)P(65 < X < 67) z1 = (65 – 65)/2.7 = 0, z2 = (67 – 65)/2.7 = 1.11z1 = (65 – 65)/2.7 = 0, z2 = (67 – 65)/2.7 = 1.11 P(65 < X < 67) = P(0 < Z < 1.11) P(65 < X < 67) = P(0 < Z < 1.11) = P(Z < 1.11) – P(Z < 0)= P(Z < 1.11) – P(Z < 0) = .867 – .5 = .367 or 36.7%= .867 – .5 = .367 or 36.7%

3.3. P(X>62) = 1-P(X>62) = 1- P(X<62) = 1-0.134 = 0.866P(X<62) = 1-0.134 = 0.866

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ExampleExample: : Suppose verbal SAT scores of high-school Suppose verbal SAT scores of high-school freshman are normally distributed with a mean of 500 freshman are normally distributed with a mean of 500 and a standard deviation of 50.and a standard deviation of 50.

What is the probability of a randomly chosen What is the probability of a randomly chosen individual having a score greater than 600?individual having a score greater than 600?

z-score = [600-500]/50 = 2z-score = [600-500]/50 = 2

P(X>600) = P(Z>2) = 1- P(Z P(X>600) = P(Z>2) = 1- P(Z 2)= 1-P(Z<2) = 2)= 1-P(Z<2) = 1-.9772 = 0.228 1-.9772 = 0.228

Note that the only difference in the two graphs below Note that the only difference in the two graphs below is the scale on the tow axes. However, the shaded is the scale on the tow axes. However, the shaded areas are equal…since the total area under any of this areas are equal…since the total area under any of this curves is one.curves is one.

300 400 500 600 700Verbal SAT Score

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300 400 500 600 700Verbal SAT Score

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P(X>600)P(X>600) P(Z>2)P(Z>2)

0 2-2

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What is the probability of a randomly chosen individual having a What is the probability of a randomly chosen individual having a score between 400 and 500?score between 400 and 500?

We want P(400<X<500).We want P(400<X<500). z-score1 = z1 = [400-500]/50 = -2z-score1 = z1 = [400-500]/50 = -2 z-score2 = z2 = [500-500]/50 = 0z-score2 = z2 = [500-500]/50 = 0 P(400<X<500) = P(-2 < Z < 0) P(400<X<500) = P(-2 < Z < 0) = P(Z<0) – P(Z<-2) = .5-.228 = .4772 (from Table) = P(Z<0) – P(Z<-2) = .5-.228 = .4772 (from Table)

-4 -2 0 2 4Z-Score

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sity

That is the probability of a randomly chosen student having a score between 400 and 500 is about .48 or 48%.

P(-2<Z<0)

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What is the probability of a randomly chosen individual having a score What is the probability of a randomly chosen individual having a score between 350 and 450?between 350 and 450?

z-score1 = z1 = [350-500]/50 = -3z-score1 = z1 = [350-500]/50 = -3 z-score2 = z2 = [450-500]/50 = -1z-score2 = z2 = [450-500]/50 = -1 P(350<X<450) = P(-3 < Z < -1) P(350<X<450) = P(-3 < Z < -1) = P(Z<-1) – P(Z<-3) = P(Z<-1) – P(Z<-3) = .1587-.0013 = .1574 (from normal Table)= .1587-.0013 = .1574 (from normal Table)

That is the probability of a randomly chosen student having a score between 350 and 450 is about .16 or 16%.

P(-3<Z<-1)

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Z-Score

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The Empirical Rule and Normal The Empirical Rule and Normal Distrib.Distrib.

The Empirical Rule states that for any bell-shaped The Empirical Rule states that for any bell-shaped distribution, approximatelydistribution, approximately 68% of the values fall within 1 standard deviation of 68% of the values fall within 1 standard deviation of

the mean in either direction. (in the interval the mean in either direction. (in the interval μμ ± ± σσ ) ) 95% of the values fall within 2 standard deviations of 95% of the values fall within 2 standard deviations of

the mean in either direction. (in the interval the mean in either direction. (in the interval μμ ± 2 ± 2σσ )) 99.7% of the values fall within 3 standard deviations 99.7% of the values fall within 3 standard deviations

of the mean in either direction. (in the interval of the mean in either direction. (in the interval μμ ± 3 ± 3σσ ))

This empirical rule is valid for all bell-shaped This empirical rule is valid for all bell-shaped distributions but it is exactly right in the case of the distributions but it is exactly right in the case of the normal distribution.normal distribution.

Check the following probabilities:Check the following probabilities:P(-1<Z<1) = ___ , P(-2<Z<2) =___ , P(-3<Z<3) =___P(-1<Z<1) = ___ , P(-2<Z<2) =___ , P(-3<Z<3) =___

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How can we find percentiles?How can we find percentiles? QuestionQuestion:: For a normal r.v. X with mean For a normal r.v. X with mean μμ and standard and standard

deviation deviation σσ , how can we find , how can we find xx (a value of (a value of X X ), such that ), such that P( X ≤ x) = α%P( X ≤ x) = α%, where , where αα is α known probability. is α known probability. e.g. if α = 95 , the 95e.g. if α = 95 , the 95thth percentile of X is the value of percentile of X is the value of XX such that such that

95% of its possible values are less than that. 95% of its possible values are less than that.

SolutionSolution: : First we get the α-th percentile for Z, First we get the α-th percentile for Z, P(Z≤ z) = 0.95 gives z = 1.64.P(Z≤ z) = 0.95 gives z = 1.64.    and we get and we get xx using using x= x= σσ z + z + μμ. .

ExampleExample:: What is the 90 What is the 90thth percentile of the height of percentile of the height of college women? [Recall that college women? [Recall that XX~N( 65, 2.7~N( 65, 2.722)])] P(Z≤ z) = 0.90, then z = 1.38 since P(Z<1.38)=0.8997 P(Z≤ z) = 0.90, then z = 1.38 since P(Z<1.38)=0.8997 the the

closest value to 0.90 in the table.closest value to 0.90 in the table. x=2.7*1.38+65=68.726, x=2.7*1.38+65=68.726, Thus the 90Thus the 90thth percentile of the height of college women is percentile of the height of college women is

68.726in. 68.726in.

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SummarySummary

Definitions and theory for Normal r.v’s.Definitions and theory for Normal r.v’s. Knowing Knowing μμ and and σσ, specifies the particular normal , specifies the particular normal

distribution out of the class of all normal distribution out of the class of all normal distributions. distributions.

The pdf of any normal r.v The pdf of any normal r.v XX~N(~N(μμ , , σσ 22)),, also called also called normal curve, is symmetric, bell shaped and normal curve, is symmetric, bell shaped and centered at centered at μμ..

The The standard normal random variable, Z,standard normal random variable, Z, has has μμ = 0 and = 0 and σσ 22 = = σσ =1. =1.

We have the tables for all the probabilities of the We have the tables for all the probabilities of the form form P(Z ≤ z) = P(Z < z). P(Z ≤ z) = P(Z < z).

For any normal r.v For any normal r.v X X ~ N(~ N(μμ , , σσ 2 2 )),, we can obtain any we can obtain any probabilities of interest using the probabilities of interest using the “standardization theorem’:“standardization theorem’:     

P(X ≤ x) = P [(X- P(X ≤ x) = P [(X- μμ)/ )/ σσ ≤ (x- ≤ (x- μμ)/ )/ σσ] = = P(Z ≤ z),] = = P(Z ≤ z),Where Where z = (x- z = (x- μμ)/ )/ σσ, , is called theis called the z-score of x. z-score of x.

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SummarySummary

Finding Probabilities of Finding Probabilities of XX~N(~N(μμ , , σσ 2 2 ))

First find the z-score of First find the z-score of xx (or (or xx’s if more than one) ’s if more than one) to be able to use the tables.to be able to use the tables.

Write the probability in terms of Write the probability in terms of Z.Z. Think what is the area under the curve that Think what is the area under the curve that

corresponds to this probability corresponds to this probability Having in mind that the normal curve is symmetric Having in mind that the normal curve is symmetric

and that the total area under the curve is equal to 1 and that the total area under the curve is equal to 1 figure out how to transform this probability into the figure out how to transform this probability into the form form P(Z<z) P(Z<z) [rules on slide ][rules on slide ]..

Finally, obtain the probability using the table.Finally, obtain the probability using the table.  

To find the α-th percentile of To find the α-th percentile of X X :: We want to find We want to find xx (a value of (a value of XX), such that ), such that P( X ≤ x) = α%P( X ≤ x) = α% First we get the α-th percentile for Z, =for example ifFirst we get the α-th percentile for Z, =for example if

α=95) α=95) P(Z≤ z) = 0.95, then z = 1.64P(Z≤ z) = 0.95, then z = 1.64 We get We get xx using using x= x= σσ z + z + μμ