Chapter 1: Statistics

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Transcript Chapter 1: Statistics

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Chapter 5: Probability Distributions (Discrete Variables)

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Chapter Goals

• Combine the ideas of frequency distributions and probability to form probability distributions.

• Investigate discrete probability distributions and study measures of central tendency and dispersion.

• Study the binomial random variable.

5.1: Random Variables

• Bridge between experimental outcomes and statistical analysis.

• Each outcome in an experiment is

assigned

to a number.

• This suggests the idea of a function.

Random Variable

: A variable that assumes a unique numerical value for each of the outcomes in the sample space of a probability experiment.

Note

: 1. Used to denote the outcomes of a probability experiment.

2. Each outcome in a probability experiment is assigned to a unique value.

3. Illustration:

S

Outcomes           Random Variable  2  1 0 1 2

Examples

of random variables: 1. Let the

number of computers sold per day

by a local merchant be a random variable. Integer values ranging from zero to about 50 are possible values.

2. Let the

number of pages in a mystery novel

at a bookstore be a random variable. The smallest number of pages is 125 while the largest number of pages is 547.

3. Let the

time it takes an employee to get to work

be a random variable. Possible values are 15 minutes to over 2 hours.

4. Let the volume of water used by a household during a month be a random variable. Amounts range up to several thousand gallons. 5. Let the number of defective components in a shipment of 1000 be a random variable. Values range from 0 to 1000.

Discrete Random Variable

: A quantitative random variable that can assume a countable number of values.

Intuitively, a discrete random variable can assume values corresponding to isolated points along a line interval. That is, there is a gap between any two values.

Note

: Usually associated with counting.

Continuous Random Variable

: A quantitative random variable that can assume an uncountable number of values.

Intuitively, a continuous random variable can assume any value along a line interval, including every possible value between any two values.

Note

: Usually associated with a measurement.

Example

: Determine whether the following random variables are discrete or continuous.

1. The barometric pressure at 12:00 PM.

2. The length of time it takes to complete a statistics exam.

3. The number of items in the shopping cart of the person in front of you at the checkout line.

4. The weight of a home grown zucchini.

5. The number of tickets issued by the PA State Police during a 24 hour period.

6. The number of cans of soda pop dispensed by a machine placed in the Mathematics building on campus.

7. The number of cavities the dentist discovers during your next visit.

5.2 Probability Distributions of a Discrete Random Variable

• Need a complete description of a discrete random variable.

• This includes all the values the random variable may assume and all of the associated probabilities.

• This information may be presented in a variety of ways.

Probability Distribution

: A distribution of the probabilities associated with each of the values of a random variable. The probability distribution is a theoretical distribution; it is used to represent populations.

Note

: 1. The probability distribution tells you everything you need to know about the random variable.

2. The probability distribution may be presented in the form of a table, chart, function, etc.

Probability Function

: A rule that assigns probabilities to the values of the random variable.

Example

: The number of people staying in a randomly selected room at a local hotel is a random variable ranging in value from 0 to 4. The probability distribution is known and is given in various forms below.

x P

(

x

) 0 2/15 1 4/15 2 5/15 3 3/15 4 1/15

Note

: 1. This chart implies the only values

x

takes on are 0, 1, 2, 3, and 4.

2.

P

( ) 

P

 5 15

A line representation of the Hotel Room probability distribution:

0 1 2 3 4

x

A

histogram

may be used to present a probability distribution.

A histogram for the Hotel Room probability distribution

: 0 1 2 3 4

x

Note

: 1. The histogram of a probability distribution uses the physical area of each bar to represent its assigned probability.

2. In the Hotel Room probability distribution: the width of each bar is 1, so the height of each bar is equal to the assigned probability, which is the area of each bar.

3. The idea of area representing probability is important in the study of continuous random variables.

Reminder

: Every probability function must satisfy the two basic properties of probability.

1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive: 0 

P

(

x

)  1 2. The sum of the probabilities assigned to all the values of the random variable must equal 1:  all

x P

(

x

)  1

5.3: Mean and Variance of a Discrete Probability Distribution

Describe the center and spread of a population.

m , s , s 2 :

population parameters

.

Population parameters are usually unknown values (we would like to estimate).

Note

: 1.

x

is the

mean of the sample

.

2.

s

2 and

s

are the

variance

and

standard deviation of the sample

.

3. ,

s

2 , and

s

are called

sample statistics

.

4.

m (lowercase Greek letter “

mu

”) is the

mean of the

5.

6.

7.

population

.

s 2 s (“

sigma squared

”) is the

variance of the population

(lowercase Greek letter “

sigma

”) is the

standard

.

deviation of the population

.

m , s 2 , and s are called

population parameters

. (A parameter is a constant. m , s 2 , and s are typically unknown values.)

Mean of a Discrete Random Variable

: The mean, m , of a discrete random variable

x

is found by multiplying each possible value of

x

by its own probability and then adding all the products together.

m   [

Note

: 1. The mean is the

average

value of the random variable, what happens on average.

2. The mean is not necessarily a value of the random variable.

Variance of a Discrete Random Variable

: Variance, s 2 , of a discrete random variable

x

is found by multiplying each possible value of the squared deviation from the mean, (

x

 m ) 2 , by its own probability and then adding all the products together.

s 2     [(

x

 m ) 2  [  [ 2 ( )]  ( )]    [ m 2  2

Standard Deviation of a Discrete Random Variable

: The positive square root of the variance.

s  s 2

Example

: The number of standby passengers who get seats on a daily commuter flight from Boston to New York is a random variable,

x

, with probability distribution given below (in an extensions table). Find the mean, variance, and standard deviation.

Totals x

0 1 2 3 4 5  0.30

0.25

0.20

0.15

0.05

0.05

1.00

(check) 0.00

0.25

0.40

0.45

0.20

0.25

1.55

 [

x

2 0 1 4 9 16 25 0.00

0.25

0.80

1.35

0.80

1.25

4.45

 [ 2

x P x

Solution

: Using the formulas for mean, variance, and standard deviation: m   [ 

Note

: 1.55 is

not

a value of the random variable (in this case). It is only what happens on average.

s 2    [ 2 ( )]    [  2   2  s  s 2  

Example

: The probability distribution for a random variable

x

is given by the probability function  8  15

x

for

x

 3, 4, 5, 6, 7 Find the mean, variance, and standard deviation.

Solution

: Find the probability associated with each value by using the probability function.

P

( )  15  5 15

P

( )  15  2 15

P

( )  15  4 15

P

( )  8  7 15  1 15

P

( )  15  3 15

Use an extensions table to find the population parameters.

Totals x

3 4 5 6 7 5/15 4/15 3/15 2/15 1/15 15/15 15/15 16/15 15/15 12/15 7/15  [ 65/15

x

2 9 16 25 36 49 2 ( ) 45/15 64/15 75/15 72/15 49/15 305/15  [ 2

x P x

m   [  65  15 s 2   [ 2 ( )]    [  2  305 15 65 15   2  s  s 2  

5.4: The Binomial Probability Distribution

• One of the most important discrete distributions.

• Based on a series of repeated trials whose outcomes can be classified in one of two categories:

success

or

failure

.

• Distribution based on a

binomial probability experiment

.

Binomial Probability Experiment

: An experiment that is made up of repeated trials that possess the following properties: 1. There are

n

repeated independent trials.

2. Each

trial

has two possible outcomes (success, failure).

3.

P

(success) =

p

,

P

(failure) =

q

, and

p

+

q

= 1 4. The

binomial random variable

x

is the count of the number of successful trials that occur;

x

may take on any integer value from zero to

n

.

Note

: 1. Properties 1 and 2 are the two basic properties of any binomial experiment.

2. Property 3 concerns the algebraic notation for each trial.

3. Property 4 concerns the algebraic notation for the complete experiment.

4. Both

x

and

p

must be associated with “success.” 5. Independent trials mean that the result of one trial does not affect the probability of success of any other trial in the experiment. The probability of “success” remains constant throughout the entire experiment.

Example

: It is known that 40% of all graduating seniors on campus have taken a statistics class. Five seniors are selected at random and asked if they have taken a statistics class.

1. A trial is asking one student, repeated 5 times. The trials are independent since the probability of taking a statistics class for any one student is not affected by the results from any other student.

2. Two outcomes on each trial: taken a statistics class (success), not taken a statistics class (failure) 3.

p

=

P

(taken a statistics class) = .40

q

=

P

(not taken a statistics class) = .60

4.

x

= number of students who have taken a statistics class

Binomial Probability Function

: For a binomial experiment, let

p

“success” and

q

represent the probability of a represent the probability of a “failure” on a single trial; then

P

(

x

), the probability that there will be exactly

x

successes on

n

trials is   

n

x

 (

p x

)(

q

), for

x

 0, 1, 2, ... , or

n Note

: 1. The number of ways that exactly

x

successes can occur in

n

trials: 

n

 

x

 2. The probability of exactly

x

successes:

p x

3. The probability that failure will occur on the remaining (

n

-

x

) trials:

q n

-

x

Note

: The number of ways that exactly

x

successes can occur in a set of

n

trials is represented by the symbol 

n

 

x

 1. Must always be a positive integer.

2. Called the

binomial coefficient

.

3. Found by using the formula: 

n

 

x

 

n

!

x

)!

n

! is an abbreviation for

n factorial.

n

!

  1 )(

n

 2 )  6 !

720

Example

: According to a recent study, 65% of all homes in a certain county have high levels of radon gas leaking into their basements. Four homes are selected at random and tested for radon. The random variable

x

is the number of homes with high levels of radon (out of the four).

Properties

: 1. There are 4 repeated trials:

n

= 4. The trials are independent.

2. Each test for radon is a trial, and each test has two outcomes:

radon

or

no radon

.

3.

p

=

P

(radon) = .65,

q

=

P

(no radon) = .35

p

+

q

= 1 4.

x

is the number of homes with high levels of radon, possible values: 0, 1, 2, 3, 4

The binomial probability function

:    4

x

  (.

65

x

) (.

35 ) 4 

x

, for

x

 0, 1, 2, 3, 4

P

( )    4 0   0 (.

65 ) (.

35 ) 4 

P

( )   4   1  (.

65 ) (.

35 ) 3  )( .

)  ) 

P

( )    4 2   (.

65 ) (.

35 ) 2 

P

( )    4 3   (.

65 ) (.

35 ) 1  )( .

)( .

)  ) 

P

( )    4 4   (.

65 ) (.

35 ) 0  )( ) 

Example

: In a certain automobile dealership, 70% of all customers purchase an extended warranty with their new car. For 15 customers selected at random: 1. Find the probability that exactly 12 will purchase an extended warranty.

2. Find the probability at most 13 will purchase an extended warranty.

Solution

: Let

x

be the number of customers who purchase an extended warranty.

x

is a binomial random variable.

The probability function associated with this experiment:   15  

x

 7

x

3 15 

x

, for

x

 0, 1, 2, ... ,15

Probability exactly 12 purchase an extended warranty:

P

( 12 )   15   12  12 3  .

1700 Probability at most 13 purchase an extended warranty:

P

(

x

 13 )   1

P

( 0 )  

P

( 1 ) 

P

( 14 )  

P

  ( 15 ) 

P

( 13 )  1       15 14   (.

7 ) 14 (.

3 ) 1  1  [.

0305  .

0047 ]    15 15   (.

7 ) 15 (.

3 ) 0    1  .

0352  .

9648

Note

: 1. The value of many binomial probabilities (

n

< 15 and common values of

p

) are found in Table 2, Appendix B.

2. Minitab has special commands for computing binomial probabilities or cumulative probabilities.

PDF 12; Binomial 15 .7.

Probability Density Function

Binomial with n = 15 and p = 0.700000

x P( X = x) 12.00 0.1700

3. Many graphing calculators also have built-in functions for computing binomial probabilities and cumulative probabilities.

Notation

: If

x

is a binomial random variable with

n

trials and probability of a success

p

, this is often denoted:

x

~ B(

n

,

p

).

Example

: Suppose

x

is a binomial random variable with

n

= 18 and

p

= .35. A convenient notation to identify this random variable is:

x

~ B(18, .35).

5.5: Mean and Standard Deviation of the Binomial Distribution

• Population parameters of the binomial distribution help to describe the distribution.

• Mean and standard deviation indicate where the distribution is centered and the spread of the distribution.

The

mean

and

standard deviation

of a theoretical binomial distribution can be found by using the following two formulas: m 

np

s 

npq Note

: 1. Mean is intuitive: number of trials multiplied by the probability of a success.

2. The variance of a binomial probability distribution is: s 2  

npq

 2 

npq

Example

: Find the mean and standard deviation of the binomial distribution when

n

= 18 and

p

= .75.

Solution

:

n

= 18,

p

= .75,

q

= 1  .75 = .25

m 

np

 ( )(.

)  s 

npq

 ( )(.

)(.

)  The probability function is   18  

x

 (.

75  25 ) 18 

x

for

x

 0, 1, 2, ... , 18

Table of values and probabilities

: x P( X = x) 0.00 0.0000

1.00 0.0000

2.00 0.0000

3.00 0.0000

4.00 0.0000

5.00 0.0000

6.00 0.0002

7.00 0.0010

8.00 0.0042

9.00 0.0139

10.00 0.0376

11.00 0.0820

12.00 0.1436

13.00 0.1988

14.00 0.2130

15.00 0.1704

16.00 0.0958

17.00 0.0338

18.00 0.0056

Histogram

: m s 1 1

x