Download ppt - Larson ch 4 Stats

Transcript
Page 1: Larson ch 4 Stats

Chapter 4

Discrete Probability Distributions

1Larson/Farber 4th ed

Page 2: Larson ch 4 Stats

Chapter Outline

• 4.1 Probability Distributions

• 4.2 Binomial Distributions• 4.3 More Discrete Probability Distributions (not in syllabus)

Larson/Farber 4th ed 2

Page 3: Larson ch 4 Stats

Section 4.1

Probability Distributions

3Larson/Farber 4th ed

Page 4: Larson ch 4 Stats

Section 4.1 Objectives

• Distinguish between discrete random variables and continuous random variables

• Construct a discrete probability distribution and its graph

• Determine if a distribution is a probability distribution

• Find the mean, variance, and standard deviation of a discrete probability distribution

• Find the expected value of a discrete probability distribution

Larson/Farber 4th ed 4

Page 5: Larson ch 4 Stats

Random Variables

Random Variable

• Represents a numerical value associated with each outcome of a probability distribution.

• Denoted by x

• Examples x = Number of sales calls a salesperson makes in

one day. x = Hours spent on sales calls in one day.

5Larson/Farber 4th ed

Page 6: Larson ch 4 Stats

Random Variables

Discrete Random Variable

• Has a finite or countable number of possible outcomes that can be listed.

• Example x = Number of sales calls a salesperson makes in

one day. x

1 530 2 4

6Larson/Farber 4th ed

Page 7: Larson ch 4 Stats

Random Variables

Continuous Random Variable

• Has an uncountable number of possible outcomes, represented by an interval on the number line.

• Example x = Hours spent on sales calls in one day.

7Larson/Farber 4th ed

x

1 2430 2 …

Page 8: Larson ch 4 Stats

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Discrete random variable (The number of stocks whose share price increases can be counted.)

x

1 3030 2 …

8Larson/Farber 4th ed

1. x = The number of stocks in the Dow Jones Industrial Average that have share price increaseson a given day.

Page 9: Larson ch 4 Stats

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces)

x

1 3230 2 …

9Larson/Farber 4th ed

2. x = The volume of water in a 32-ouncecontainer.

Page 10: Larson ch 4 Stats

Textbook Exercises. Page 197

• Distinguish between Discrete and Continuous random variable

14. x represents the length of time it takes to go to work.

x is a continuous random variable because length of time is a measurement and not a count. Measurements are continuous.

16. x represents number of tornadoes in the month of June in Oklahoma.

x is a discrete random variable because number of tornadoes is a count. Counts are discrete.

Larson/Farber 4th ed 10

Page 11: Larson ch 4 Stats

Discrete Probability Distributions

Discrete probability distribution

• Lists each possible value the random variable can assume, together with its probability.

• Must satisfy the following conditions:

11Larson/Farber 4th ed

In Words In Symbols1. The probability of each value of

the discrete random variable is between 0 and 1, inclusive.

2. The sum of all the probabilities is 1.

0 P (x) 1

ΣP (x) = 1

Page 12: Larson ch 4 Stats

What is a PROBABILITY?

0% 25% 50% 75% 100%

0 ¼ or .25 ½ 0r .5 ¾ or .75 1

Impossible Not Very Equally Likely Somewhat Certain

Likely Likely

Larson/Farber 4th ed 12

Page 13: Larson ch 4 Stats

Constructing a Discrete Probability Distribution

1. Make a frequency distribution for the possible outcomes.

2. Find the sum of the frequencies.

3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.

4. Check that each probability is between 0 and 1 and that the sum is 1.

13Larson/Farber 4th ed

Let x be a discrete random variable with possible outcomes x1, x2, … , xn.

Page 14: Larson ch 4 Stats

Example: Constructing a Discrete Probability Distribution

An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely

Score, x Frequency, f

1 24

2 33

3 42

4 30

5 21

aggressive. A score of 3 indicated neither trait. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram.

14Larson/Farber 4th ed

Page 15: Larson ch 4 Stats

Solution: Constructing a Discrete Probability Distribution

• Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable.

24(1) 0.16

150P

33(2) 0.22

150P 42

(3) 0.28150

P

30(4) 0.20

150P

21(5) 0.14

150P

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14

• Discrete probability distribution:

15Larson/Farber 4th ed

Page 16: Larson ch 4 Stats

Solution: Constructing a Discrete Probability Distribution

This is a valid discrete probability distribution since

1. Each probability is between 0 and 1, inclusive,0 ≤ P(x) ≤ 1.

2. The sum of the probabilities equals 1, ΣP(x) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14

16Larson/Farber 4th ed

Page 17: Larson ch 4 Stats

Solution: Constructing a Discrete Probability Distribution

• Histogram

Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome.

17Larson/Farber 4th ed

Page 18: Larson ch 4 Stats

Textbook Exercises. Page 198

• Problem 22 Blood Donations

x being number of donations is countable and hence a discrete random variable.

Also, number of donations are mutually exclusive.

a. P(X>1) = P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6)

= 0.25 + 0.10 + 0.05 + 0.03 + 0.02 = 0.45

b. P(X < 3) = P(x = 2) + P(x = 1) + P(x = 0)

= 0.25 + 0.25 + 0.30 = 0.80

• Problem 24 Missing Probability

Since it is a probability distribution sum of all the probabilities is equal to one.

missing value = 1 – sum of rest of the probabilities

= 1 – 0.85 = 0.15

Larson/Farber 4th ed 18

x 0 1 2 3 4 5 6

P(x) 0.05 ? 0.23 0.21 0.17 0.11 0.08

Page 19: Larson ch 4 Stats

Mean

Mean of a discrete probability distribution

• μ = ΣxP(x)

• Each value of x is multiplied by its corresponding probability and the products are added.

19Larson/Farber 4th ed

Page 20: Larson ch 4 Stats

x P(x) xP(x)

1 0.16 1(0.16) = 0.16

2 0.22 2(0.22) = 0.44

3 0.28 3(0.28) = 0.84

4 0.20 4(0.20) = 0.80

5 0.14 5(0.14) = 0.70

Example: Finding the Mean

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean.

μ = ΣxP(x) = 2.94

Solution:

20Larson/Farber 4th ed

Page 21: Larson ch 4 Stats

Variance and Standard Deviation

Variance of a discrete probability distribution

• σ2 = Σ(x – μ)2P(x)

Standard deviation of a discrete probability distribution

2 2( ) ( )x P x

21Larson/Farber 4th ed

Page 22: Larson ch 4 Stats

Example: Finding the Variance and Standard Deviation

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ = 2.94)

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14

22Larson/Farber 4th ed

Page 23: Larson ch 4 Stats

Solution: Finding the Variance and Standard Deviation

Recall μ = 2.94

x P(x) x – μ (x – μ)2 (x – μ)2P(x)

1 0.16 1 – 2.94 = –1.94 (–1.94)2 = 3.764 3.764(0.16) = 0.602

2 0.22 2 – 2.94 = –0.94 (–0.94)2 = 0.884 0.884(0.22) = 0.194

3 0.28 3 – 2.94 = 0.06 (0.06)2 = 0.004 0.004(0.28) = 0.001

4 0.20 4 – 2.94 = 1.06 (1.06)2 = 1.124 1.124(0.20) = 0.225

5 0.14 5 – 2.94 = 2.06 (2.06)2 = 4.244 4.244(0.14) = 0.594

2 1.616 1.3 Standard Deviation:

Variance:

23Larson/Farber 4th ed

σ2 = Σ(x – μ)2P(x) = 1.616

Page 24: Larson ch 4 Stats

Textbook Exercises. Page 199

• Problem 30 Camping Chairs

Construct the probability distribution and determine :

a. Mean b. Standard Deviation c. Variance

Larson/Farber 4th ed 24

Defects, x Batches, f P(x)

0 95 95 ÷ 380 = 0.250

1 113 113 ÷ 380 = 0.297

2 87 0.229

3 64 0.168

4 13 0.034

5 8 0.021

n = 380 ∑ P(x) = 1

To determine the mean, standard deviation and variance we will use TI 83/84. First store the x-values and the P(x) values in two of the lists, say L1 and L2. Make sure that x goes into L1 and P(x) goes into L2. Do not reverse that order. Then using 1-var stats from CALC menu enter L1 comma L2 and hit the ENTER key. You may notice that the sample standard deviation is not shown but you may use population standard deviation to get an estimate. Double check your results with hand calculations.

1.54 variancesample

1.24deviation standard sample

5.1xmean sample

2

Page 25: Larson ch 4 Stats

Expected Value

Expected value of a discrete random variable

• Equal to the mean of the random variable.

• E(x) = μ = ΣxP(x)

25Larson/Farber 4th ed

Page 26: Larson ch 4 Stats

Example: Finding an Expected Value

At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket. What is the expected value of your gain?

26Larson/Farber 4th ed

Page 27: Larson ch 4 Stats

Solution: Finding an Expected Value

• To find the gain for each prize, subtractthe price of the ticket from the prize: Your gain for the $500 prize is $500 – $2 = $498 Your gain for the $250 prize is $250 – $2 = $248 Your gain for the $150 prize is $150 – $2 = $148 Your gain for the $75 prize is $75 – $2 = $73

• If you do not win a prize, your gain is $0 – $2 = –$2

27Larson/Farber 4th ed

Page 28: Larson ch 4 Stats

Solution: Finding an Expected Value

• Probability distribution for the possible gains (outcomes)

Gain, x $498 $248 $148 $73 –$2

P(x)1

1500

1

1500

1

1500

1

1500

1496

1500

( ) ( )

1 1 1 1 1496$498 $248 $148 $73 ( $2)1500 1500 1500 1500 1500$1.35

E x xP x

You can expect to lose an average of $1.35 for each ticket you buy.

28Larson/Farber 4th ed

Page 29: Larson ch 4 Stats

Section 4.1 Summary

• Distinguished between discrete random variables and continuous random variables

• Constructed a discrete probability distribution and its graph

• Determined if a distribution is a probability distribution

• Found the mean, variance, and standard deviation of a discrete probability distribution

• Found the expected value of a discrete probability distribution

Larson/Farber 4th ed 29

Page 30: Larson ch 4 Stats

Section 4.2

Binomial Distributions

Larson/Farber 4th ed 30

Page 31: Larson ch 4 Stats

Section 4.2 Objectives

• Determine if a probability experiment is a binomial experiment

• Find binomial probabilities using the binomial probability formula

• Find binomial probabilities using technology and a binomial table

• Graph a binomial distribution

• Find the mean, variance, and standard deviation of a binomial probability distribution

Larson/Farber 4th ed 31

Page 32: Larson ch 4 Stats

Binomial Experiments

1. The experiment is repeated for a fixed number of trials, where each trial is independent of other trials.

2. There are only two possible outcomes of interest for each trial. The outcomes can be classified as a success (S) or as a failure (F).

3. The probability of a success P(S) is the same for each trial.

4. The random variable x counts the number of successful trials.

Larson/Farber 4th ed 32

Page 33: Larson ch 4 Stats

Notation for Binomial Experiments

Symbol Description

n The number of times a trial is repeated

p = P(s) The probability of success in a single trial

q = P(F) The probability of failure in a single trial (q = 1 – p)

x The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, … , n.

Larson/Farber 4th ed 33

Page 34: Larson ch 4 Stats

Example: Binomial Experiments

Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x.

Larson/Farber 4th ed 34

1. A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on eight patients. The random variable represents the number of successful surgeries.

Page 35: Larson ch 4 Stats

Solution: Binomial Experiments

Binomial Experiment

1. Each surgery represents a trial. There are eight surgeries, and each one is independent of the others.

2. There are only two possible outcomes of interest for each surgery: a success (S) or a failure (F).

3. The probability of a success, P(S), is 0.85 for each surgery.

4. The random variable x counts the number of successful surgeries.

Larson/Farber 4th ed 35

Page 36: Larson ch 4 Stats

Solution: Binomial Experiments

Binomial Experiment

• n = 8 (number of trials)

• p = 0.85 (probability of success)

• q = 1 – p = 1 – 0.85 = 0.15 (probability of failure)

• x = 0, 1, 2, 3, 4, 5, 6, 7, 8 (number of successful surgeries)

Larson/Farber 4th ed 36

Page 37: Larson ch 4 Stats

Example: Binomial Experiments

Decide whether the experiment is a binomial experiment. If it is, specify the values of n, p, and q, and list the possible values of the random variable x.

Larson/Farber 4th ed 37

2. A jar contains five red marbles, nine blue marbles, and six green marbles. You randomly select three marbles from the jar, without replacement. The random variable represents the number of red marbles.

Page 38: Larson ch 4 Stats

Solution: Binomial Experiments

Not a Binomial Experiment

• The probability of selecting a red marble on the first trial is 5/20.

• Because the marble is not replaced, the probability of success (red) for subsequent trials is no longer 5/20.

• The trials are not independent and the probability of a success is not the same for each trial.

Larson/Farber 4th ed 38

Page 39: Larson ch 4 Stats

Textbook Exercises. Pages 211 - 212

• Problem 4 Graphical Analysis

a) p = 0.75 (graph is skewed to left implies p is greater than 0.5 )

b) p = 0.50 (graph is symmetric)

c) p = 0.25 (graph is skewed to right implies p is smaller than 0.5)

• Problem 10 Clothing store purchases

Since it satisfies all the requirements , it is a binomial experiment.

where, Success is: person does not make a purchase

n = 18, p = 0.74, q = 0.26 and x = 0, 1,2, …, 18

• Problem 12 Lottery

It is not a binomial experiment because probability of success is not same for each trial.

Larson/Farber 4th ed 39

Page 40: Larson ch 4 Stats

Binomial Probability Formula

40Larson/Farber 4th ed

Binomial Probability Formula

• The probability of exactly x successes in n trials is

!( )

( )! !x n x x n x

n x

nP x C p q p q

n x x

• n = number of trials

• p = probability of success

• q = 1 – p probability of failure

• x = number of successes in n trials

Page 41: Larson ch 4 Stats

Example: Finding Binomial Probabilities

Microfracture knee surgery has a 75% chance of success on patients with degenerative knees. The surgery is performed on three patients. Find the probability of the surgery being successful on exactly two patients.

Larson/Farber 4th ed 41

Page 42: Larson ch 4 Stats

Solution: Finding Binomial Probabilities

Method 1: Draw a tree diagram and use the Multiplication Rule

Larson/Farber 4th ed 42

9(2 ) 3 0.422

64P successful surgeries

Page 43: Larson ch 4 Stats

Solution: Finding Binomial Probabilities

Method 2: Binomial Probability Formula

Larson/Farber 4th ed 43

2 3 2

3 2

2 1

3 1(2 )

4 4

3! 3 1

(3 2)!2! 4 4

9 1 273 0.422

16 4 64

P successful surgeries C

3 13, , 1 , 2

4 4n p q p x

Page 44: Larson ch 4 Stats

Binomial Probability Distribution

Binomial Probability Distribution

• List the possible values of x with the corresponding probability of each.

• Example: Binomial probability distribution for Microfacture knee surgery: n = 3, p =

Use binomial probability formula to find probabilities.

Larson/Farber 4th ed 44

x 0 1 2 3

P(x) 0.016 0.141 0.422 0.422

3

4

Page 45: Larson ch 4 Stats

Example: Constructing a Binomial Distribution

In a survey, workers in the U.S. were asked to name their expected sources of retirement income. Seven workers who participated in the survey are randomly selected and asked whether they expect to rely on Social

Larson/Farber 4th ed 45

Security for retirement income. Create a binomial probability distribution for the number of workers who respond yes.

Page 46: Larson ch 4 Stats

Solution: Constructing a Binomial Distribution

• 25% of working Americans expect to rely on Social Security for retirement income.

• n = 7, p = 0.25, q = 0.75, x = 0, 1, 2, 3, 4, 5, 6, 7

Larson/Farber 4th ed 46

P(x = 0) = 7C0(0.25)0(0.75)7 = 1(0.25)0(0.75)7 ≈ 0.1335

P(x = 1) = 7C1(0.25)1(0.75)6 = 7(0.25)1(0.75)6 ≈ 0.3115

P(x = 2) = 7C2(0.25)2(0.75)5 = 21(0.25)2(0.75)5 ≈ 0.3115

P(x = 3) = 7C3(0.25)3(0.75)4 = 35(0.25)3(0.75)4 ≈ 0.1730

P(x = 4) = 7C4(0.25)4(0.75)3 = 35(0.25)4(0.75)3 ≈ 0.0577

P(x = 5) = 7C5(0.25)5(0.75)2 = 21(0.25)5(0.75)2 ≈ 0.0115

P(x = 6) = 7C6(0.25)6(0.75)1 = 7(0.25)6(0.75)1 ≈ 0.0013

P(x = 7) = 7C7(0.25)7(0.75)0 = 1(0.25)7(0.75)0 ≈ 0.0001

Page 47: Larson ch 4 Stats

Solution: Constructing a Binomial Distribution

Larson/Farber 4th ed 47

x P(x)

0 0.1335

1 0.3115

2 0.3115

3 0.1730

4 0.0577

5 0.0115

6 0.0013

7 0.0001

All of the probabilities are between 0 and 1 and the sum of the probabilities is 1.00001 ≈ 1.

Page 48: Larson ch 4 Stats

Example: Finding Binomial Probabilities

A survey indicates that 41% of women in the U.S. consider reading their favorite leisure-time activity. You randomly select four U.S. women and ask them if reading is their favorite leisure-time activity. Find the probability that at least two of them respond yes.

Larson/Farber 4th ed 48

Solution: • n = 4, p = 0.41, q = 0.59• At least two means two or more.• Find the sum of P(2), P(3), and P(4).

Page 49: Larson ch 4 Stats

Solution: Finding Binomial Probabilities

Larson/Farber 4th ed 49

P(x = 2) = 4C2(0.41)2(0.59)2 = 6(0.41)2(0.59)2 ≈ 0.351094

P(x = 3) = 4C3(0.41)3(0.59)1 = 4(0.41)3(0.59)1 ≈ 0.162654

P(x = 4) = 4C4(0.41)4(0.59)0 = 1(0.41)4(0.59)0 ≈ 0.028258

P(x ≥ 2) = P(2) + P(3) + P(4) ≈ 0.351094 + 0.162654 + 0.028258 ≈ 0.542

Page 50: Larson ch 4 Stats

Example: Finding Binomial Probabilities Using Technology

The results of a recent survey indicate that when grilling, 59% of households in the United States use a gas grill. If you randomly select 100 households, what is the probability that exactly 65 households use a gas grill? Use a technology tool to find the probability. (Source: Greenfield Online for Weber-Stephens Products Company)

Larson/Farber 4th ed 50

Solution:• Binomial with n = 100, p = 0.59, x = 65

Page 51: Larson ch 4 Stats

Solution: Finding Binomial Probabilities Using Technology

Larson/Farber 4th ed 51

From the displays, you can see that the probability that exactly 65 households use a gas grill is about 0.04.

Page 52: Larson ch 4 Stats

Example: Finding Binomial Probabilities Using a Table

About thirty percent of working adults spend less than 15 minutes each way commuting to their jobs. You randomly select six working adults. What is the probability that exactly three of them spend less than 15 minutes each way commuting to work? Use a table to find the probability. (Source: U.S. Census Bureau)

Larson/Farber 4th ed 52

Solution:• Binomial with n = 6, p = 0.30, x = 3

Page 53: Larson ch 4 Stats

Solution: Finding Binomial Probabilities Using a Table

• A portion of Table 2 is shown

Larson/Farber 4th ed 53

The probability that exactly three of the six workers spend less than 15 minutes each way commuting to work is 0.185.

Page 54: Larson ch 4 Stats

Example: Graphing a Binomial Distribution

Fifty-nine percent of households in the U.S. subscribe to cable TV. You randomly select six households and ask each if they subscribe to cable TV. Construct a probability distribution for the random variable x. Then graph the distribution. (Source: Kagan Research, LLC)

Larson/Farber 4th ed 54

Solution: • n = 6, p = 0.59, q = 0.41• Find the probability for each value of x

Page 55: Larson ch 4 Stats

Solution: Graphing a Binomial Distribution

Larson/Farber 4th ed 55

x 0 1 2 3 4 5 6

P(x) 0.005 0.041 0.148 0.283 0.306 0.176 0.042

Histogram:

Page 56: Larson ch 4 Stats

Textbook Exercises. Page 213

• Problem 22 Honeymoon Financing

n = 20 married couples p = 0.70 (probability of success)

Follow the procedure described bellow to find the probabilities using TI 83/84

a. P(x = 1)

Press 2nd key and DISTR to see the list of different distributions. Scroll down until you see binompdf(. Choose that option and then enter the values of n comma p comma x in this order and close the parenthesis. Hit ENTER.

binompdf(20, 0.7, 1) 1.627 10 -9

b. P(x 1) = 1 – [P(x = 0) + P( x = 1)] (using complement rule)

= 1 – [ binompdf(20, 0.7,0) + binompdf(20 , 0.7, 1)]

= 1 – [ 3.487 10-11 + 1.627 10-9 ] = 0.99999 1

c. P (x 1) = 1 – P(x 1) (using complement rule)

= 1 – 0.9999

0

Larson/Farber 4th ed 56

Page 57: Larson ch 4 Stats

Textbook Exercises. Page 213

• Problem 26 Movies on Phone

n = 12 adults p = 0.25 (probability of success)

Follow the procedure described bellow to find the probabilities using TI 83/84

a. P(x = 4)

Press 2nd key and DISTR to see the list of different distributions. Scroll down until you see binompdf(. Choose that option and then enter the values of n comma p comma x in this order and close the parenthesis. Hit ENTER.

binompdf(12, 0.25, 4) = 0.194

b. P(x 4) = 1 – P(x 4) (using complement rule)

= 1 – [P(0) + P(1) + P(2) + P(3) + P(4)]

= 1 – [0.032 + 0.127 + 0.232 + 0.258 + 0.194 ] = 0.157

c. P (4 x 8) = P(4) + P(5) + P(6) + P(7) + P(8)

= 0.194 + 0.103+ 0.04 + 0.011 + 0.002

= 0.35

Larson/Farber 4th ed 57

Page 58: Larson ch 4 Stats

Mean, Variance, and Standard Deviation

• Mean: μ = np

• Variance: σ2 = npq

• Standard Deviation:

Larson/Farber 4th ed 58

npq

Page 59: Larson ch 4 Stats

Example: Finding the Mean, Variance, and Standard Deviation

In Pittsburgh, Pennsylvania, about 56% of the days in a year are cloudy. Find the mean, variance, and standard deviation for the number of cloudy days during the month of June. Interpret the results and determine any unusual values. (Source: National Climatic Data Center)

Larson/Farber 4th ed 59

Solution: n = 30, p = 0.56, q = 0.44

Mean: μ = np = 30∙0.56 = 16.8Variance: σ2 = npq = 30∙0.56∙0.44 ≈ 7.4Standard Deviation: 30 0.56 0.44 2.7npq

Page 60: Larson ch 4 Stats

Solution: Finding the Mean, Variance, and Standard Deviation

Larson/Farber 4th ed 60

μ = 16.8 σ2 ≈ 7.4 σ ≈ 2.7

• On average, there are 16.8 cloudy days during the month of June.

• The standard deviation is about 2.7 days. • Values that are more than two standard deviations

from the mean are considered unusual. 16.8 – 2(2.7) =11.4, A June with 11 cloudy days

would be unusual. 16.8 + 2(2.7) = 22.2, A June with 23 cloudy

days would also be unusual.

Page 61: Larson ch 4 Stats

Section 4.2 Summary

• Determined if a probability experiment is a binomial experiment

• Found binomial probabilities using the binomial probability formula

• Found binomial probabilities using technology and a binomial table

• Graphed a binomial distribution

• Found the mean, variance, and standard deviation of a binomial probability distribution

Larson/Farber 4th ed 61


Recommended