1 Chapter 6 Continuous Probability Distributions.

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Chapter 6

Continuous Probability Distributions

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Chapter Outline

Uniform Probability Distribution Normal Probability Distribution

f (x)f (x)

x x

UniformUniform

xx

f f ((xx)) NormalNormal

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A continuous random variable assume values that have no gap or jump between them.

Since between any two values, a continuous random variable assumes infinite number of values, the probability that any particular value occurs is zero.

Therefore, we study the probability of the random variable assuming a value within a given interval.

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The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the line of the probability density function between x1 and x2.

f (x)f (x)

x x

UniformUniform

xx11 xx11 xx22 xx22

xx

f f ((xx)) NormalNormal

xx11 xx11 xx22 xx22

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Uniform Probability Distribution

A random variable is uniformly distributed whenever the A random variable is uniformly distributed whenever the probability is proportional to the interval’s length.probability is proportional to the interval’s length.

The uniform probability density function is:The uniform probability density function is:

f f ((xx) = 1/() = 1/(bb – – aa) for ) for aa << xx << bb = 0 elsewhere= 0 elsewhere f f ((xx) = 1/() = 1/(bb – – aa) for ) for aa << xx << bb = 0 elsewhere= 0 elsewhere

where: where: aa = smallest value the variable can assume = smallest value the variable can assume

bb = largest value the variable can assume = largest value the variable can assume

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Var(Var(xx) = () = (bb - - aa))22/12/12Var(Var(xx) = () = (bb - - aa))22/12/12

E(E(xx) = () = (aa + + bb)/2)/2E(E(xx) = () = (aa + + bb)/2)/2

Expected Value of Expected Value of xx

Variance of Variance of xx

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Example: Slater's BuffetExample: Slater's Buffet

Slater customers are charged for the amount ofSlater customers are charged for the amount ofsalad they take. Sampling suggests that the amountsalad they take. Sampling suggests that the amountof salad taken is uniformly distributed between 5of salad taken is uniformly distributed between 5ounces and 15 ounces.ounces and 15 ounces.

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Example: Slater's BuffetExample: Slater's Buffet Uniform Probability Density FunctionUniform Probability Density Function

ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 15 15

= 0 elsewhere= 0 elsewhere

ff((xx) = 1/10 for 5 ) = 1/10 for 5 << xx << 15 15

= 0 elsewhere= 0 elsewhere

where:where:

xx = salad plate filling weight = salad plate filling weight

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Expected Value of Expected Value of xx

E(E(xx) = () = (aa + + bb)/2)/2

= (5 + 15)/2= (5 + 15)/2

= 10= 10

E(E(xx) = () = (aa + + bb)/2)/2

= (5 + 15)/2= (5 + 15)/2

= 10= 10

Variance of Variance of xx

Var(Var(xx) = () = (bb - - aa))22/12/12

= (15 – 5)= (15 – 5)22/12/12

= 8.33= 8.33

Var(Var(xx) = () = (bb - - aa))22/12/12

= (15 – 5)= (15 – 5)22/12/12

= 8.33= 8.33

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Uniform Probability Distribution Uniform Probability Distribution for Salad Plate Filling

Weight

x x

1/101/10

Salad Weight (oz.)Salad Weight (oz.)

55 1010 151500

f(x)f(x)

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What is the probability that a customer will take between 12 and 15 ounces of salad?

x x

1/101/10

Salad Weight (oz.)Salad Weight (oz.)

55 1010 151500

f(x)f(x)

P(12 < x < 15) = (1/10)(3) = .3P(12 < x < 15) = (1/10)(3) = .3

1212

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Area as A Measure of Probability

The area under the graph of The area under the graph of ff((xx) is simply the ) is simply the measure of probability.measure of probability.

This is valid for all continuous random This is valid for all continuous random variables.variables. The probability that The probability that xx takes on a value takes on a value between some lower value between some lower value xx11 and some higher and some higher value value xx22 can be found by computing the area can be found by computing the area under the graph of under the graph of ff((xx) over the interval from ) over the interval from xx11 to to xx22. . The overall area under the graph of The overall area under the graph of ff((xx) over ) over the value range of the value range of xx is 1 or 100%. is 1 or 100%.

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Normal Probability Distribution

The normal probability distribution is the most important The normal probability distribution is the most important distribution for describing a continuous random variable.distribution for describing a continuous random variable.

It is widely used in statistical inference.It is widely used in statistical inference. It has been used in describing a wide variety of real-life It has been used in describing a wide variety of real-life

applications including:applications including:

• Heights of Heights of peoplepeople• Rainfall Rainfall amountsamounts

• Test scoresTest scores• Scientific Scientific measurementsmeasurements

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Normal Probability Density FunctionNormal Probability Density Function

2 2( ) / 21( )

2xf x e

2 2( ) / 21( )

2xf x e

= mean= mean

= standard deviation= standard deviation

= 3.14159= 3.14159

ee = 2.71828 = 2.71828

where:where:

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CharacteristicsCharacteristics

The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero. The distribution is The distribution is symmetricsymmetric; its skewness; its skewness measure is zero.measure is zero.

xx

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The entire family of normal probability The entire family of normal probability distributions is defined by its mean distributions is defined by its mean and its and its standard deviation standard deviation ..

The entire family of normal probability The entire family of normal probability distributions is defined by its mean distributions is defined by its mean and its and its standard deviation standard deviation ..

xxMean

Standard Deviation Standard Deviation

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The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the median and modemedian and mode.. The The highest pointhighest point on the normal curve is at the on the normal curve is at the meanmean, which is also the , which is also the median and modemedian and mode..

xxMean

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The mean can be any numerical value. When only The mean can be any numerical value. When only the mean (the central location) changes, the whole the mean (the central location) changes, the whole normal curve simply shifts horizontally. normal curve simply shifts horizontally.

The mean can be any numerical value. When only The mean can be any numerical value. When only the mean (the central location) changes, the whole the mean (the central location) changes, the whole normal curve simply shifts horizontally. normal curve simply shifts horizontally.

xx-8-8 00 2424

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The standard deviation determines the width of theThe standard deviation determines the width of thecurve. A smaller curve. A smaller results in a narrower, taller curve. results in a narrower, taller curve. The standard deviation determines the width of theThe standard deviation determines the width of thecurve. A smaller curve. A smaller results in a narrower, taller curve. results in a narrower, taller curve.

xx

= 5

= 12

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Normal Probability Distribution CharacteristicsCharacteristics

Probabilities for the normal random variable are Probabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area under. The total area undercurve is 1 (.5 to the left of the mean andcurve is 1 (.5 to the left of the mean and .5 to the right..5 to the right.

Probabilities for the normal random variable are Probabilities for the normal random variable are given by given by areas under the curveareas under the curve. The total area under. The total area undercurve is 1 (.5 to the left of the mean andcurve is 1 (.5 to the left of the mean and .5 to the right..5 to the right.

xxMean

.5.5 .5.5

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Characteristics (basis for the empirical rule)

Expected number of correct answers

of values of a normal random variableof values of a normal random variable are between are between - - and and + + .. of values of a normal random variableof values of a normal random variable are between are between - - and and + + ..68.26%68.26%68.26%68.26%

of values of a normal random variableof values of a normal random variable are between are between - 2 - 2 and and + 2 + 2.. of values of a normal random variableof values of a normal random variable are between are between - 2 - 2 and and + 2 + 2..95.44%95.44%95.44%95.44%

of values of a normal random variableof values of a normal random variable are between are between - 3 - 3 and and + 3 + 3.. of values of a normal random variableof values of a normal random variable are between are between - 3 - 3 and and + 3 + 3..99.72%99.72%99.72%99.72%

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xx – – 33 – – 11

– – 22 + 1+ 1

+ 2+ 2 + 3+ 3

68.26%68.26%95.44%95.44%99.72%99.72%

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Standard Normal Probability Distribution



A normally distributed random variable A normally distributed random variable with a mean of 0 and a standard deviation of 1 iswith a mean of 0 and a standard deviation of 1 is said to have a said to have a standard normal probabilitystandard normal probability distributiondistribution. Any other normal distribution can be . Any other normal distribution can be converted to the standard normal distribution.converted to the standard normal distribution.

A normally distributed random variable A normally distributed random variable with a mean of 0 and a standard deviation of 1 iswith a mean of 0 and a standard deviation of 1 is said to have a said to have a standard normal probabilitystandard normal probability distributiondistribution. Any other normal distribution can be . Any other normal distribution can be converted to the standard normal distribution.converted to the standard normal distribution.

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Characteristics


The letter z is used to designate the stand normal The letter z is used to designate the stand normal random variable.random variable. The letter z is used to designate the stand normal The letter z is used to designate the stand normal random variable.random variable.

zz0

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Converting to the Standard Normal Distribution

0

x

z

We can think of We can think of zz as a measure of the number of as a measure of the number ofstandard deviations standard deviations xx is from is from ..

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Example: What is the probability that z 1.58?

0 1.58z

Area of z 1.58 is the probability.

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Example: What is the probability that z 1.58? Cumulative Probability Table for the Standard Normal Cumulative Probability Table for the Standard Normal

Distribution: it provides the areas (probabilities) to the Distribution: it provides the areas (probabilities) to the LEFTLEFT of of

the the zz values. values.z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

. . . . . . . . . . .

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .94411.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .95451.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .96331.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .97061.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

. . . . . . . . . . .P(z 1.58)

=.9429

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Example: What is the probability that z 1.58?

0 1.58z

Area = .9429 Area = 1-.9429 = .0571

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Example: Computer Prices

The average price of personal computers manufactured The average price of personal computers manufactured by Company APO is $1,000 with a standard deviation of by Company APO is $1,000 with a standard deviation of $150. Furthermore, it is known that the computer prices $150. Furthermore, it is known that the computer prices manufactured by APO are normally distributed. manufactured by APO are normally distributed.

aa. What is the probability that a randomly selected . What is the probability that a randomly selected computer will have a price of at least $1125?computer will have a price of at least $1125?

PP((xx 1125) = ? 1125) = ?

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Step 1: Convert Step 1: Convert xx to the standard normal distribution. to the standard normal distribution.Step 1: Convert Step 1: Convert xx to the standard normal distribution. to the standard normal distribution.

zz = ( = (xx - - )/)/ = (1125 – 1000)/150= (1125 – 1000)/150 = .83= .83

zz = ( = (xx - - )/)/ = (1125 – 1000)/150= (1125 – 1000)/150 = .83= .83

Step 2: Find the area under the standard normalStep 2: Find the area under the standard normal curve to the left of curve to the left of zz = .83. = .83.Step 2: Find the area under the standard normalStep 2: Find the area under the standard normal curve to the left of curve to the left of zz = .83. = .83.

see next slidesee next slide see next slidesee next slide

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z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

. . . . . . . . . . .

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

. . . . . . . . . . .


Example: Computer Prices Cumulative Probability Table for the Standard Normal Cumulative Probability Table for the Standard Normal

Distribution:Distribution:

PP((zz << .83) .83)

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Step 3: Compute the area under the standard normalStep 3: Compute the area under the standard normal curve to the right of curve to the right of zz = .83. = .83.Step 3: Compute the area under the standard normalStep 3: Compute the area under the standard normal curve to the right of curve to the right of zz = .83. = .83.

PP((z z .83) = 1 – .83) = 1 – PP((zz << .83) .83) = 1- .7967= 1- .7967

= .2033= .2033

PP((z z .83) = 1 – .83) = 1 – PP((zz << .83) .83) = 1- .7967= 1- .7967

= .2033= .2033

PP((xx 1125)1125)

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00 .83.83

Area = .7967Area = .7967Area = 1 - .7967Area = 1 - .7967

= .2033= .2033

zz

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bb. To attract buyers like college students, Company APO . To attract buyers like college students, Company APO decides to provide promotions on the decides to provide promotions on the cheapercheaper computers. What is the computers. What is the maximummaximum price such that only price such that only 15% of all the computers are eligible for a promotion 15% of all the computers are eligible for a promotion sale? sale?

--------------------------------------------------------------------------------------------------------------------------------------------(Hint: Given a probability (area), we can use (Hint: Given a probability (area), we can use

the standard normal table in an inverse the standard normal table in an inverse fashion to find the corresponding fashion to find the corresponding zz value.) value.)

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z

Area = .15Area = .15Area = .15Area = .15

0 zz.85.85zz.15.15

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Step 1: Determine whether the z value is negative or positive. The question is asking for the maximum computer price such that the cheapest computers (15% of all the computers) can enjoy a promotion sale. It is translated as: P(x <<pricepricemaxmax) = .15) = .15 Therefore, the z value should be negative and it cuts off an area of .15 in the left tail of the standard normal distribution.

Step 1: Determine whether the z value is negative or positive. The question is asking for the maximum computer price such that the cheapest computers (15% of all the computers) can enjoy a promotion sale. It is translated as: P(x <<pricepricemaxmax) = .15) = .15 Therefore, the z value should be negative and it cuts off an area of .15 in the left tail of the standard normal distribution.

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Step 2: Find the z-value. The following table shows the positive z-values. To find the negative z-values, we can utilize the symmetric feature of the Normal Distributions.

Step 2: Find the z-value. The following table shows the positive z-values. To find the negative z-values, we can utilize the symmetric feature of the Normal Distributions.

z

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

1 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621

1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830

1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015

1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177

1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319The closest z-value that cuts off the left area of .85 is 1.04. Therefore, the z-value we look for is –1.04.

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Step 3: Convert z.15 to the corresponding value of x. Step 3: Convert z.15 to the corresponding value of x.

xx = = + + zz.15.15

= 1000 + (-1.04)(150)= 1000 + (-1.04)(150)

= 844= 844

xx = = + + zz.15.15

= 1000 + (-1.04)(150)= 1000 + (-1.04)(150)

= 844= 844

So, the maximum price at which a computer can enjoySo, the maximum price at which a computer can enjoya promotion sale is $844, and there are about 15% of all a promotion sale is $844, and there are about 15% of all the computers manufactured by APO are eligible for thethe computers manufactured by APO are eligible for thepromotion sale.promotion sale.

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z

Area = .15Area = .15

1000844

= 150= 150

1 Chapter 6 Continuous Probability Distributions.

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Transcript of 1 Chapter 6 Continuous Probability Distributions.