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Statistics with Economics and Statistics with Economics and Business ApplicationsBusiness Applications

Chapter 5 The Normal and Other Continuous Probability Distributions

Normal Probability Distribution

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ReviewReview

I. I. What’s in last lecture?What’s in last lecture?

Binomial, Poisson and Hypergeometric Probability Distributions. Chapter 4.

II. II. What's in this lecture?What's in this lecture?

Normal Probability Distribution. Read Chapter 5

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Continuous Random VariablesContinuous Random Variables

A random variable is continuous if it can assume the infinitely many values corresponding to points on a line interval.

• Examples:Examples:

– Heights, weights

– length of life of a particular product

– experimental laboratory error

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Continuous Probability DistributionContinuous Probability Distribution Suppose we measure height of students in this class. If we

“discretize” by rounding to the nearest feet, the discrete probability histogram is shown on the left. Now if height is measured to the nearest inch, a possible probability histogram is shown in the middle. We get more bins and much smoother appearance. Imagine we continue in this way to measure height more and more finely, the resulting probability histograms approach a smooth curve shown on the right.

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Probability Distribution for a Continuous Probability Distribution for a Continuous Random VariableRandom Variable

Probability distribution describes how the probabilities are distributed over all possible values. A probability distribution for a continuous random variable x is specified by a mathematical function denoted by f(x) which is called the density function. The graph of a density function is a smooth curve.

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Properties of Continuous Probability Properties of Continuous Probability DistributionsDistributions

• f(x) 0• The area under the curve is equal to 11..• P(a x b) = area under the curvearea under the curve between a and

b.

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Some IllustrationsSome Illustrations

a

P(x<a)

Notice that for a continuous random variable x, P(x = a) = 0 for any specific value a because the “area above a point” under the curve is a line segment and hence has 0 area. Specifically this means P(x<a) = P(x a) P(a<x<b) = P(ax<b) = P(a<xb) = P(a xb)

b

P(x>b)

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Method of Probability CalculationMethod of Probability CalculationThe probability that a continuous random variable x lies between a lower limit a and an upper limit b is

P(a<x<b) = (cumulative area to the left of b) –

(cumulative area to the left of a)

= P(x < b) – P(x < a)

= -a b ab

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Continuous Probability Continuous Probability DistributionsDistributions

• There are many different types of continuous random variables

• We try to pick a model that– Fits the data well– Allows us to make the best possible

inferences using the data.• One important continuous random variable is

the normal random variablenormal random variable.

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The Normal DistributionThe Normal Distribution

deviation. standard andmean population theare and

1416.3 7183.2

for 2

1)(

2

2

1

e

xexfx

deviation. standard andmean population theare and

1416.3 7183.2

for 2

1)(

2

2

1

e

xexfx

Two parameters, mean and standard deviation, completely determine the Normal distribution. The shape and location of the normal curve changes as the mean and standard deviation change.

The formula that generates the normal probability distribution is:

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Normal Distributions: =1

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

0, 1

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

1, 1

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

2, 1

Normal Distributions: =1

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

3, 1

0.00

0.10

0.20

0.30

0.40

0.50

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

1, 1

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Normal Distributions: Normal Distributions: =0=0

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

=0, =1

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

=0, =1=0, =2

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

=0, =1=0, =2=0, =3

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

=0, =1=0, =2=0, =3=0, =0.5

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

1.60

1.80

-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0

=0, =1=0, =2=0, =3=0, =0.5=0, =0.25

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The Standard Normal The Standard Normal DistributionDistribution

• To find P(a < x < b), we need to find the area under the appropriate normal curve.

• To simplify the tabulation of these areas, we standardizestandardize each value of x by expressing it as a z-score, the number of standard deviations it lies from the mean .

x

z

x

z

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The Standard The Standard Normal (Normal (zz) )

DistributionDistribution

• Mean = 0; Standard deviation = 1• When x = , z = 0• Symmetric about z = 0• Values of z to the left of center are negative• Values of z to the right of center are positive• Total area under the curve is 1.• Areas on both sides of center equal .5

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Using Table 3Using Table 3The four digit probability in a particular row and column of Table 3 gives the area under the standard normal curve between 0 and a positive value z. This is enough because the standard normal curve is symmetric.

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To find an area between 0 and a positive z-value, read directly from the tableUse properties of standard normal curve and other probability rules to find other areas

To find an area between 0 and a positive z-value, read directly from the tableUse properties of standard normal curve and other probability rules to find other areas

Using Table 3Using Table 3

P(0<z<1.96) = .4750

P(-1.96<z<0)= P(0<z<1.96)=.4750

P(z<1.96)=P(z<0)+ P(0<z<1.96)=.5+.4750=.9750

P(z<-1.96)=P(z>1.96)=.5-.4750=.0250

P(-1.96<z<1.96)=P(z<1.96)-P(z<-1.96)

=.9750-.0250=.9500

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1. Look for the four digit area closest to .4750 in Table 3.

2. What row and column does this value correspond to?

1. Look for the four digit area closest to .4750 in Table 3.

2. What row and column does this value correspond to?

Working BackwardsWorking Backwards

Often we know the area and want to find the z-value that gives the area.

Example: Find the value of a positive z that has area .4750 between 0 and z.

3. z = 1.96

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ExampleExample

P(z<?) = .75

P(z<?)=P(z<0)+P(0<z<?)=.5+P(0<z<?)=.75

P(0<z<?)=.25

z = .67

What percentile does this value represent?

75th percentile, or the third quartile.

What percentile does this value represent?

75th percentile, or the third quartile.

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1. The area to its left will be 1 - .05 = .95

2. The area to its left and right to 0 will be .95-.5=.45

3. Look for the four digit area closest to .4500 in Table 3.

4. Since the value .4500 is halfway between .4495 and .4505, we choose z halfway between 1.64 and 1.65. z=1.645

1. The area to its left will be 1 - .05 = .95

2. The area to its left and right to 0 will be .95-.5=.45

3. Look for the four digit area closest to .4500 in Table 3.

4. Since the value .4500 is halfway between .4495 and .4505, we choose z halfway between 1.64 and 1.65. z=1.645

Working BackwardsWorking Backwards

Find the value of z that has area .05 to its right.

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Finding Probabilities for the Finding Probabilities for the General Normal Random VariableGeneral Normal Random Variable

To find an area for a normal random variable x with mean and standard deviation , standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

To find an area for a normal random variable x with mean and standard deviation , standardize or rescale the interval in terms of z. Find the appropriate area using Table 3.

Example: Example: x has a normal distribution with mean = 5 and sd = 2. Find P(x > 7).

1587.3413.5.1)10()0(1

)1(1)1()2

57()7(

zPzP

zPzPzPxP

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ExampleExampleThe weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation .10. What is the probability that a randomly selected package weighs between 0.80 and 0.85 pounds?

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)1.

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1.

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zPzP

zPzP

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ExampleExampleWhat is the weight of a package such that only 5% of all packages exceed this weight?

16.11)1(.645.1?

645.11.

1? 3, Table From

45.50.95.)1.

1?0(

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ExampleExample

A Company produces “20 ounce” jars of a picante sauce. The true amounts of sauce in the jars of this brand sauce follow a normal distribution.

Suppose the companies “20 ounce” jars follow a normally distribution with a mean =20.2 ounces with a standard deviation =0.125 ounces.

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Example Example What proportion of the jars are under-filled (i.e.,

have less than 20 ounces of sauce)?

P(z<-1.60) = P(z>1.60) = P(z>0)-P(0<z<1.60) = .5-.4452 = .0548. The proportion of the sauce jars that are under-filled is .0548

60.1125.0

2.2020

xz

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What proportion of the sauce jars contain between 20 and 20.3 ounces of sauce.

Z

20 2020125

160.

.. Z

203 2020125

080. ..

.

P(-1.60<z<.80) = P(-1.60<z<0)+P(0<z<.80) = P(0<z<1.60)+P(0<z<.80)=.4452+.2881=.7333

P(-1.60<z<.80) = P(z<.80)-P(z<-1.60)=.5+P(0<z<.80)-[.5-P(0<z<1.60)]=P(0<z<1.60)+P(0<z<.80)=.7333

Example Example

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99% of the jars of this brand of picante sauce will contain more than what amount of sauce?

Example Example

91.19)125(.33.22.20?

33.2125.

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49.)125.

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33.2125.

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How Probabilities Are DistributedHow Probabilities Are Distributed

• The interval contains approximately 68% of the measurements.

• The interval 2 contains approximately 95% of the measurements.

• The interval 3 contains approximately 99.7% of the measurements.

• The interval contains approximately 68% of the measurements.

• The interval 2 contains approximately 95% of the measurements.

• The interval 3 contains approximately 99.7% of the measurements.

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The Normal Approximation to the The Normal Approximation to the BinomialBinomial

• We can calculate binomial probabilities using– The binomial formula– The cumulative binomial tables

• When n is large, and p is not too close to zero or one, areas under the normal curve with mean np and variance npq can be used to approximate binomial probabilities.

SticiGuiSticiGui

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Approximating the BinomialApproximating the Binomial

Make sure to include the entire rectangle for the values of x in the interval of interest. This is called the continuity correctioncontinuity correction. . Standardize the values of x using

npqnp,σ,μσ

μxz

npqnp,σ,μ

σ

μxz

Make sure that np and nq are both greater than 5 to avoid inaccurate approximations! Orn is large and 2 falls between 0 and n (book)

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Correction for ContinuityCorrection for Continuity Add or subtract .5 to include the entire rectangle. For

illustration, suppose x is a Binomial random variable with n=6, p=.5. We want to compute P(x 2). Using 2 directly will miss the green area. P(x 2)=P(x 2.5) and use 2.5.

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ExampleExampleSuppose x is a binomial random variable with n = 30 and p = .4. Using the normal approximation to find P(x 10).

n = 30 p = .4 q = .6

np = 12 nq = 18

683.2)6)(.4(.30

12)4(.30

Calculate

npq

np

683.2)6)(.4(.30

12)4(.30

Calculate

npq

np

The normal approximation

is ok!

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ExampleExample

)683.2

125.10()10(

zPxP

2877.)56.( zP

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ExampleExample

)xP()xP(

)xP()xP(

P(x)P(x

)xP)P(x

xP)P(x

5.95.4105

5.95.5105

)5.55

5.4(5

)5.9(10

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ExampleExampleA production line produces AA batteries with a reliability rate of 95%. A sample of n = 200 batteries is selected. Find the probability that at least 195 of the batteries work.

Success = working battery n = 200

p = .95 np = 190 nq = 10

The normal approximation

is ok!

))05)(.95(.200

1905.194()195(

zPxP

0722.9278.1)46.1( zP

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Key ConceptsKey ConceptsI. Continuous Probability DistributionsI. Continuous Probability Distributions

1. Continuous random variables

2. Probability distributions or probability density functions

a. Curves are smooth.

b. The area under the curve between a and b represents

the probability that x falls between a and b.

c. P (x a) 0 for continuous random variables.

II. The Normal Probability DistributionII. The Normal Probability Distribution

1. Symmetric about its mean .

2. Shape determined by its standard deviation .

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Key ConceptsKey ConceptsIII. The Standard Normal DistributionIII. The Standard Normal Distribution

1. The normal random variable z has mean 0 and standard deviation 1.2. Any normal random variable x can be transformed to a standard normal random variable using

3. Convert necessary values of x to z.4. Use Table 3 in Appendix I to compute standard normal probabilities.5. Several important z-values have tail areas as follows:

Tail Area: .005 .01 .025 .05 .10

z-Value: 2.58 2.33 1.96 1.645 1.28

x

z

x

z