Ch 6 Key

Chapter 6 Discrete Probability Distributions

Ch6.1 Discrete Random Variables

Objective A: Discrete Probability Distribution

A1. Distinguish between Discrete and Continuous Random Variables

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.

Discreten = 0, 1, 2, 3 …

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

Continuousd > 0

A2. Discrete Probability Distributions

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

Table (a)

Not a discrete probability distribution because it does not meet ∑ P (x) = 1.

Table (b)

It is a discrete probability distribution because it meets ∑ P (x) = 1

and each P(x) is between 0 and 1.

Example 2: (a) Determine the required value of the missing probability to make the distribution

a discrete probability distribution.

1

(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) Draw a probability histogram.

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

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

(1)

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

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

Formula (2a) in the textbook

X*P(x)

0

0.073

1

0.117

2

0.258

3

0.322

4

0.230

*Use StatCrunch to find the mean and standard deviation of the discrete random variable in Example 1.

Steps: 1) Enter the data: X and P(X).

2) Click Stat → Calculators → Custom.

3) Choose X for Values in: and P(X) for Weights in:

4) Click Compute!

Objective C : Expected Value

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

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)

X * P (x)

Gain

+350

0.998734

350 * (0.998734) = 349.5569

Loss

-249650

1-0.998734 = 0.001266

-249650*(0.001266) = -316.0569

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

Chapter 6.2 The 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.

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 repondents 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.

Objective B : Binomial Formula

Let the random variablebe the number of successes in trials of a binomial experiment.

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

(Round to four decimal places as needed)

P (x)

P (x = 12)

= 455 ≈ 0.2184

Example 2: (a) Use StatCrunch to compute a Binomial table of .

First, state the possible values of the random variable , then Open StatCrunch select Stat Calculators Binomial Standard Input For P(x = 0), select = in the inequality box Input 0 Compute Record the result Repeat for each x value 1, 2, 3 and 4 to complete the table.

x

0

1

2

3

4

x

0

0.01500625

1

0.111475

2

0.3105375

3

0.384475

4

0.17850625

(b) Use the Binomial table from part (a), find .P(x > 2) = 0.384475 + 0.17850625 = 0.56298125

(c) Use the Binomial table from part (a), find .

= 0.01500625 + 0.111475 + 0.3105375 = 0.43701875

Objective C : Binomial Table by StatCunch

Example 1: Use StatCrunch with Binomial Distribution to find with and . Open StatCrunch select Stat Calculators Binomial Standard Input n = 12 and p = 0.4 For P(x ≤ 6), select in the inequality box Compute and record the result. (If you need to include the graph, right click on image, select copy image, Paste Special, then select Device Independent Bitmap)

Example 2According to the American Lung Association, 90% of adult smokers started smoking beforeturning 21 years old. Ten smokers 21 years old or older are randomly selected,and thenumber 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 is the same for each trial of experiment

−The trials are independent

− Fixed numbers of trials n = 10

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

Open StatCrunch Stat Calculator Binomial Standard Input n = 10 and p = 0.9 For P(x = 8), select = in the inequality box Compute and record the result.

The probability that exactly 8 started smoking before 21 years of age is

P (x = 8) = 0.19371024

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

Open StatCrunch Stat Calculator Binomial Standard Input n = 10 and

p = 0.9 For P(x ≥ 8) , select in the inequality box Compute and record the result.

P ( x = 8 or more) = P (x ≥ 8) = 0.92980917

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

Open StatCrunch Stat Calculator Binomial Between Input n = 10 and p = 0.9 For P(7 ≤ x ≤ 9), Input 7 and 9 in the compound inequality box Compute

and record the result.

P (7 ≤ X ≤ 9) = 0.63852636

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

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

= n p = 9 * 0.8 = 7.2

= np(1- p)= 9(0.8)(1-0.8) = 9(0.8)(0.2) = 1.2

Example 2: According to the 2005 American Community Survey, 43% of women aged 18 to 24 wereenrolled in college in 2005.

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

= n*p = 500 *0.43 = 215

= 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.

What role do and play in the shape of a binomial distribution? Study the textbook pg. 318-319.

In Other Words

Provided that the interval to represent the “usual” observations. Observations outside this interval may be considered unusual.

)

230

.

0

(

4

×

481

.

1

519

.

2

4

=

-

50447303

.

0

)

230

.

0

(

)

481

.

1

(

2

=

18

.

1

381639

.

1

»

=

x

s

381639

.

1

)]

(

)

[(

2

2

=

×

-

=

å

x

P

x

x

x

m

s

)

(

)

(

2

x

P

x

x

×

-

m

x

x

m

-

[

]

()()

ExxPx

=×

å

x

x

5

.

33

)]

(

[

)

(

=

×

=

å

x

P

x

x

E

x

n

x

n

15,0.85,12

npx

===

4 and 0.65

np

==

x

()

Px

å

=

73

.

0

)

(

x

P

(2)

Px

>

(03)

Px

£<

(6)

Px

£

12

n

=

0.4

p

=

£

³

x

90.8

np

==

x

43

.

0

,

500

=

=

p

n

p

n

(1)10,

npp

-³

2

ms

-

2

ms

+

å

=

1

)

(

x

P

50

.

0

40

.

0

30

.

0

20

.

0

10

.

0

0

1

2

3

4

5

)

(

x

P

x

50

.

0

40

.

0

30

.

0

20

.

0

10

.

0

0

1

2

3

4

5

)

(

x

P

x

x

[()]

x

xPx

m

=×

å

x

()

Px

()

xPx

×

519

.

2

=

å

×

=

)]

(

[

x

P

x

x

m

22

[()()]

xx

xPx

sm

=-×

å

)

073

.

0

(

0

×

519

.

2

519

.

2

0

-

=

-

463211353

.

0

)

073

.

0

(

)

519

.

2

(

2

=

-

)

117

.

0

(

1

×

519

.

1

519

.

2

1

-

=

-

069495138

.

0

)

258

.

0

(

)

519

.

0

(

2

=

-

269961237

.

0

)

117

.

0

(

)

519

.

1

(

2

=

-

)

258

.

0

(

2

×

519

.

0

519

.

2

2

-

=

-

)

322

.

0

(

3

×

481

.

0

519

.

2

3

=

-

074498242

.

0

)

322

.

0

(

)

481

.

0

(

2

=

]

)

(

[

å

×

=

x

P

x

x

m