Ch 6.1 to 6.2 Los Angeles Mission College Produced by DW Copyright of the definitions and examples...

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Ch 6.1 to 6.2

Los Angeles Mission College

Copyright of the definitions and examples is reserved to Pearson Education, Inc.. In order to use this PowerPoint presentation, the required textbook for the class is the Fundamentals of Statistics, Informed Decisions Using Data, Michael Sullivan, III, fourth edition.

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Chapter 6.1 Discreet Random VariablesObjective A : Discrete Probability Distribution

Objective B : Mean and Standard Deviation of a Discrete Random Variable

Chapter 6.2 Binomial Probability Distribution

Objective A : Criteria for a Binomial Probability Experiment

Objective B : Binomial Formula

Objective C : Binomial Table

Objective C : Mean Expected Value

Objective D : Mean and Standard Deviation of a Binomial Random VariableObjective E : Applications


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Chapter 6.1 Discreet Random VariablesObjective A : Discrete Probability DistributionA1. Distinguish between Discrete and Continuous Random Variables


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Example 1: Determine whether the random variable is discrete or continuous. State the possible values of the random variable.

(a) The number of fish caught during the fishing tournament.

(b) The distance of a baseball travels in the air after being hit.

Continuous

Discrete 0, 1, 2, 3,...n

0d


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A2. Discrete Probability Distributions


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0 0.34

1 0.21

2 0.13

3 0.04

4 0.01

Example 1: Determine whether the distribution is a discrete probability distribution. If not, state why.

(a) x )(xP

Not a discreet probability distribution because it does not meet . 1)(xP

73.0)(xP


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0 0. 40

1 0.31

2 0.23

3 0.04

4 0.02

(b)

It is a discreet probability distribution because it meets . 1)(xP

x )(xP

1)(xP


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0 0. 30

1 0.15

2 ?

3 0.20

4 0.15

5 0.05

Example 2 : (a) Determine the required value of the missing probability to make the distribution a discrete probability distribution.(b) Draw a probability histogram.

x )(xP


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1)5()4()3()2()1()0( xPxPxPxPxPxP105.015.020.0)2(15.030.0 xP

(a) The required value of the missing probability

(b) The probability histogram

185.0)2( xP85.01)2( xP

15.0

50.0

40.0

30.0

20.0

10.0

0 1 2 3 4 5

)(xP

x


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Chapter 6.1 Discreet Random Variables


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0

1

2

3

4

258.0

322.0

230.0

073.0

117.0

)(xPx

(a) Mean

0)073.0(0 117.0)117.0(1

516.0)258.0(2

966.0)322.0(3 920.0)230.0(4

)(xPx

)]([ xPxx

519.2

)]([ xPxx (1)

Example 1: Find the mean, variance, and standard deviation of the discrete random variable.


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0

1

2

3

4

258.0

322.0

230.0

073.0

117.0

)(xPx519.2519.20

519.1519.21

519.0519.22

481.0519.23

481.1519.24

xx

18.1381639.1 x

(b) Variance ---> Use the definition formula

463211353.0)073.0()519.2( 2

269961237.0)117.0()519.1( 2 069495138.0)258.0()519.0( 2

074498242.0)322.0()481.0( 2 50447303.0)230.0()481.1( 2

)()( 2 xPx x

381639.1)]()[( 22 xPx xx

)073.0(0

)117.0(1

)258.0(2

)322.0(3

)230.0(4

)(xPx

])([ xPxx

519.2

])()[( 22 xPx xx (2a)


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0

1

2

3

4

258.0

322.0

230.0

073.0

117.0

)(xPx

222 )]([ xx xPx

(c) Variance ---> Use the computation formula

22 519.2727.7 x

381639.1x18.1

0)073.0(02 117.0)117.0(12 032.1)258.0(22

898.2)322.0(32 68.3)230.0(42

)(2 xPx

727.7)]([ 2 xPx

381639.12x

)073.0(0

)117.0(1

)258.0(2

)322.0(3)230.0(4

)(xPx

519.2])([ xPxx

222 )]([ xx xPx (2b)


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The mean of a random variable is the expected value, , of the probability experiment in the long run. In game theory is positive for money gained and is negative for money lost.

)]([)( xPxxEx x

Chapter 6.1 Discreet Random Variables


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Example 1: A life insurance company sells a $250,000 1-year term life insurance policy to a 20-year-old male for $350. According to the National Vital Statistics Report, 56(9), the probability that the male survives the year is 0.998734. Compute and interpret the expected value of this policy to the insurance company.

)(xP

998734.0

x

Gain

Loss

In the long run, the insurance company will profit $33.50 per 20-year-old male.

)(xPx

5569.349)998734.0(350

0569.316)001266.0(249650

5.33)]([)( xPxxE

350

249650

Gain/Loss

998734.01001266.0


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Example 2: Shawn and Maddie purchase a foreclosed property for $50,000 and spend an additional $27,000 fixing up the property. They feel that they can resell the property for $120,000 with probability 0.15, $100,000 with probability

0.45, $80,000 with probability 0.25, and $60,000 with probability 0.15. Compute and interpret the expected profit for reselling the property.

)(xP

15.0

0

45.0

25.0

15.0

x000,77000,27000,50

000,43000,77000,120

000,17000,77000,60

000,23000,77000,100 000,3000,77000,80

In the long run, the expected gain is $15,000 per house.

)(xPx

450,6)15.0(000,43 0)0(000,77

350,10)45.0(23000 750)25.0(000,3

550,2)15.0(000,17

Gain/Loss

Loss

Loss

Gain

Gain

Gain

000,15)]([)( xPxxE


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Chapter 6.2 Binomial Probability DistributionObjective A : Criteria for a Binomial Probability Experiment

The binomial probability distribution is a discrete probability distribution that obtained from a binomial experiment.


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Example 1: Determine which of the following probability experiments represents a binomial experiment. If the probability experiment is not a binomial experiment, state why.

Not a binomial distribution because the mileage can have more than 2 outcomes.

(b) A poll of 1,200 registered voters is conducted in which the respondents are asked whether they believe Congress should reform Social Security.

A binomial distribution because – there are 2 outcomes. (should or should not reform Social Security)– fixed number of trials. (n = 1200)– the trials are independent.– we assume the probability of success is the same for each trial of experiment.

(a) A random sample of 30 cars in a used car lot is obtained, and their mileages recorded.


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Objective B : Binomial Formula Let the random variable be the number of successes in trials of a binomial experiment.

x n



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Example 1: A binomial probability experiment is conducted with the given parameters. Compute the probability of successes in the independent trials of the experiment.

xn

12,85.0,15 xpnxnx

xn ppCxP )1()(

1215121215 )85.01()85.0()12( CxP

312 )15.0()85.0(455 0.2184


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Example 2: According to the 2005 American Community Survey, 43% of

women aged 18 to 24 were enrolled in college in 2005. Twenty-five women aged 18 to 24 are randomly selected, and the number of enrolled in college is recorded.(a) Find the probability that exactly 15 of the women are enrolled in college.

15,43.0,25 xpn

xnxxn ppCxP )1()(

15 103268760(0.43) (0.57)

1525151525 )43.01()43.0()15( CxP

0376.0


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(b) Find the probability that between 11 and 13 of the women, inclusive,

are enrolled in college.

)13()12()11()1311( xPxPxPxP

1325131325

1225121225

1125111125

)43.01()43.0(

)43.01()43.0()43.01()43.0(

C

CC

1213

13121411

)57.0()43.0(5200300

)57.0()43.0(5200300)57.0()43.0(4457400

4027.0

1311,43.0,25 xpn


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Objective C : Binomial TableAnother method for obtaining binomial probabilities is the binomial probability table. Table IV in Appendix A gives cumulative probabilities of a binomial random variable such as .Table III in Appendix A gives the exact probability of a binomial random variable such as .

)6( xP

Example 1: Use the Binomial Table to find with and .

)6( xP 12n0.4p

From Cumulative Binomial Probability Distribution (Table IV), .8418.0)6( xP

From Binomial Probability Distribution (Table III), ( 6) ( 6) ( 5) ( 4) ( 3)

( 2) ( 1) ( 0)

P x P x P x P x P x

P x P x P x

8418.0

0.1766 0.2270 0.2128 0.1419 0.0639

0.0174 0.0022


x


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Example 2: According to the American Lung Association, 90% of adult smokers started smoking before turning 21 years old. Ten smokers 21 years old or older are randomly selected, and the number of smokers who started smoking before 21 is recorded.

(a) Explain why this is a binomial experiment.

– There are 2 outcomes (smoke or not)

– The probability of success trial is the same for each trial of experiment

– The trials are independent

– Fixed numbers of trials


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1937.0)8( xP

(b) Use the binomial formula to find the probability that exactly 8 of them started smoking before 21 years of age.

8,9.0,10 xpn

From Binomial Probability Distribution (Table III),

)10()9()8()8( xPxPxPxP

9298.0

(c) Use the binomial table to find the probability that at least 8 of them started smoking before 21 years of age.

8,9.0,10 xpn

3487.03874.01937.0



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)9()8()7()97( xPxPxPxP

6385.0

(d) Use the binomial table to find the probability that between 7 and 9 of them, inclusive, started smoking before 21 years of age.

97,9.0,10 xpn

3874.01937.00574.0



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Objective D : Mean and Standard Deviation of a Binomial Random Variable

8.0,9 pn2.78.09 pnx

2.1)2.0)(8.0(9)8.01)(8.0(9)1( pnpx

Example 1: A binomial probability experiment is conducted with the given parameters. Compute the mean and standard deviation of the random variable .x



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Example 2: According to the 2005 American Community Survey, 43% of

women aged 18 to 24 were enrolled in college in 2005.

43.0,500 pn

21543.0500 pnx)43.01)(43.0(500)1( pnpx

(b) Interpret the mean.

(a) For 500 randomly selected women ages 18 to 24 in 2005, compute the mean and standard deviation of the random variable , the number of women who were enrolled in college.

x

An average of 215 out of 500 randomly selected women aged 18 to 24 were enrolled in college.

070.11)57.0)(43.0(500


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(c) Of the 500 randomly selected women, find the interval that would be

considered "usual“ for the number of women who were enrolled in college.

Data fall within 2 standard deviations of the mean are considered to be usual.

xxInterval 2)07.11(2215

)14.237,86.192(

(d) Would it be unusual if 200 out of the 500 women were enrolled in college? Why?

No, because 200 is within the interval obtained in part (c). It is not unusual to find 200 out of 500 women were enrolled in college in 2005.

14.22215


Ch 6.1 to 6.2 Los Angeles Mission College Produced by DW Copyright of the definitions and examples...

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