Ch 6.1 to 6.2_09_14 (1) - Los Angeles Mission College

```Ch 6.1 to 6.2
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 Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
A1. 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.
Discrete
n  0, 1, 2, 3,...
(b) The distance of a baseball travels in the air after being hit.
Continuous
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d 0
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A2. Discrete Probability Distributions
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Example 1: Determine whether the distribution is a discrete
probability distribution. If not, state why.
(a)
x
P (x )
0
0.34
1
0.21
2
0.13
3
0.04
4
0.01
 P(x)  0.73
Not a discreet probability distribution because it does not
meet  P(x) 1 .
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(b)
x
P (x )
0
0. 40
1
0.31
2
0.23
3
0.04
4
0.02
 P(x) 1
It is a discreet probability distribution because it meets
 P(x) 1 .
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Example 2 :
(a) Determine the required value of the missing probability to make
the distribution a discrete probability distribution.
(b) Draw a probability histogram.
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x
P (x )
0
0. 30
1
0.15
2
?
3
0.20
4
0.15
5
0.05
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(a) The required value of the missing probability
P( x  0)  P( x  1)  P( x  2)  P( x  3)  P( x  4)  P( x  5)  1
0.30  0.15  P( x  2)  0.20  0.15  0.05  1
P( x  2)  0.85  1
P( x  2)  1  0.85
 0.15
(b) The probability histogram
P (x )
0.50
0.40
0.30
0.20
0.10
x
0
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1
2
3
4
5
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.1 Discreet Random Variables
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
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Example 1: Find the mean, variance, and standard deviation of the
discrete random variable.
(a) Mean
 x  [ x  P( x)]
(1)
x
P (x )
x  P (x)
0
0.073
0  (0.073)  0
1
0.117
1 (0.117)  0.117
2
0.258
2  (0.258)  0.516
3
0.322
3  (0.322)  0.966
4
0.230
4  (0.230)  0.920
 x  [ x  P( x)]
 2.519
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(b) Variance ---> Use the definition formula
 x  [( x   x ) 2  P( x)]
2
(2a)
x  x
P (x )
x  P (x)
0 0.073
0  (0.073)
1 0.117
1 (0.117)
2 0.258
2  (0.258)
1 2.519  1.519 (1.519) 2 (0.117)  0.269961237
2  2.519  0.519 (0.519) 2 (0.258)  0.069495138
3 0.322
3 (0.322)
3  2.519  0.481
(0.481) 2 (0.322)  0.074498242
4 0.230
4  (0.230)
4  2.519  1.481
(1.481) 2 (0.230)  0.50447303
x
( x   x ) 2  P( x)
2
0  2.519  2.519 (2.519) (0.073)  0.463211353
 x  [ x  P( x)]
 2.519
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 x 2  [( x   x ) 2  P( x)]  1.381639
 x  1.381639  1.18
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(c) Variance ---> Use the computation formula
 x 2  [ x 2  P( x)]   x 2
(2b)
x
P (x )
x  P (x)
0
0.073
0  (0.073)
02  (0.073)  0
1
0.117
1 (0.117)
2
0.258
2  (0.258)
12  (0.117)  0.117
22  (0.258)  1.032
3
0.322
3 (0.322)
32  (0.322)  2.898
4
0.230
4  (0.230)
42  (0.230)  3.68
 x  [ x  P( x)]
 2.519
x 2  P( x)
2
[
x
  P( x)]  7.727
 x 2  [ x 2  P( x)]   x 2
 x 2  7.727  2.5192
 x 2  1.381639
 x 1.381639
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 1.18
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.1 Discreet Random Variables
Objective C : Mean Expected Value
The mean of a random variable is the expected value,
E ( x)  [ x  P( x)] , of the probability experiment in the long
run. In game theory x is positive for money gained and x is
negative for money lost.
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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.
Gain/Loss
x
P (x )
Gain
 350
0.998734
 249650
1 0.998734
 0.001266
Loss
x  P (x)
350  (0.998734)  349.5569
 249650  (0.001266)  316.0569
E( x)  [ x  P( x)]  33.5
In the long run, the insurance company will profit \$33.50 per 20-yearold male.
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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.
Gain/Loss
x
P (x )
x  P (x)
Loss
50,000  27,000  77,000
Gain
120,000  77,000  43,000
0
0.15
 77,000  (0)  0
43,000  (0.15)  6,450
Gain
100,000  77,000  23,000
23000  (0.45)  10,350
Gain
Loss
80,000  77,000  3,000
0.45
0.25
60,000  77,000  17,000
0.15
3,000  (0.25)  750
 17,000  (0.15)  2,550
E( x)  [ x  P( x)]  15,000
In the long run, the expected gain is \$15,000 per house.
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.2 Binomial Probability Distribution
Objective 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.
(a) A random sample of 30 cars in a used car lot is obtained,
and their mileages recorded.
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.
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.2 Binomial Probability Distribution
Objective B : Binomial Formula
Let the random variable x be the number of successes in n trials
of a binomial experiment.
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Example 1: A binomial probability experiment is conducted with the
given parameters. Compute the probability xof successes
in the n independent trials of the experiment.
n  15,
p  0.85,
x  12
P( x)  n C x p x (1  p) n  x
P( x  12)  15 C12 (0.85)12 (1  0.85)1512
 455(0.85)12 (0.15)3  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.
n  25,
p  0.43,
x  15
P( x)  n C x p x (1  p) n  x
P( x  15)  25 C15 (0.43)15 (1  0.43) 2515
 3268760(0.43)15 (0.57)10
 0.0376
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(b) Find the probability that between 11 and 13 of the women, inclusive,
are enrolled in college.
n  25,
p  0.43, 11  x  13
P(11  x  13)  P( x  11)  P( x  12)  P( x  13)
 25C11 (0.43)11 (1  0.43) 2511  25C12 (0.43)12 (1  0.43) 2512
 25C13 (0.43)13 (1  0.43) 2513
 4457400 (0.43)11 (0.57)14  5200300(0.43)12 (0.57)13
 5200300(0.43)13 (0.57)12
 0.4027
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.2 Binomial Probability Distribution
Objective C : Binomial Table
Another method for obtaining binomial probabilities is the
binomial probability table. Table IV in Appendix A gives cumulative
probabilities of a binomial random variable x such as P( x  6) .
Table III in Appendix A gives the exact probability of a binomial
random variable  such as ( = 6).
Example 1: Use the Binomial Table to find P( x  6) with n  12and
p  0.4
.
From Cumulative Binomial Probability Distribution
(Table IV), P( x  6)  0.8418 .
From Binomial Probability Distribution (Table III),
P( x  6)  P( x  6)  P( x  5)  P( x  4)  P( x  3)
 P( x  2)  P( x  1)  P( x  0)
 0.1766  0.2270  0.2128  0.1419  0.0639
 0.0174  0.0022
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 0.8418
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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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(b) Use the binomial formula to find the probability that exactly 8 of
them started smoking before 21 years of age.
n  10,
p  0.9,
x 8
From Binomial Probability Distribution (Table III),
P( x  8)  0.1937
(c) Use the binomial table to find the probability that at least 8 of them
started smoking before 21 years of age.
n  10,
p  0.9,
x8
From Binomial Probability Distribution (Table III),
P( x  8)  P( x  8)  P( x  9)  P( x  10)
 0.1937  0.3874  0.3487
 0.9298
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(d) Use the binomial table to find the probability that between
7 and 9 of them, inclusive, started smoking before 21 years of age.
n  10,
p  0.9,
7 x9
From Binomial Probability Distribution (Table III),
P(7  x  9)  P( x  7)  P( x  8)  P( x  9)
 0.0574  0.1937  0.3874
 0.6385
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Chapter 6.1 Discreet Random Variables
Objective A : Discrete Probability Distribution
Objective B : Mean and Standard Deviation of a Discrete
Random Variable
Objective C : Mean Expected Value
Chapter 6.2 Binomial Probability Distribution
Objective A : Criteria for a Binomial Probability Experiment
Objective B : Binomial Formula
Objective C : Binomial Table
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Objective E : Applications
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Chapter 6.2 Binomial Probability Distribution
Objective D : Mean and Standard Deviation of a Binomial
Random Variable
Example 1: A binomial probability experiment is conducted with the
given parameters. Compute the mean and standard
deviation of the random variable x .
n  9, p  0.8
 x  n  p  9  0.8  7.2
 x np(1  p)  9(0.8)(1  0.8)  9(0.8)(0.2)  1.2
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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.
(a) For 500 randomly selected women ages 18 to 24 in 2005, compute
the mean and standard deviation of the random variable x , the
number of women who were enrolled in college.
n  500,
p  0.43
 x  n  p  500  0.43  215
 x np(1  p)  500(0.43)(1  0.43)
 500(0.43)(0.57)  11.070
(b) Interpret the mean.
An average of 215 out of 500 randomly selected women aged 18
to 24 were enrolled in college.
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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.
Interval   x  2 x
 215  2 (11.07)
 215 22.14
 (192.86, 237.14)
(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.
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