Stats Review Chapters 5-6

Created by Teri Johnson Math Coordinator, Mary Stangler Center for Academic Success

Examples are taken from Statistics 4 E by Michael Sullivan, III

And the corresponding Test Generator from Pearson

Revised 12/13

Note:

This review is composed of questions the textbook and the test generator. This review is meant to highlight basic concepts from the course. It does not cover all concepts presented by your instructor. Refer back to your notes, unit objectives, handouts, etc. to further prepare for your exam. A copy of this review can be found at www.sctcc.edu/cas.

The final answers are displayed in red and the chapter/section number is the corner.

Probability Rules 1) P(event) is greater or equal to 0 and less than or

equal to 1.

2) The sum of the probabilities must equal 1.

Is the table a Probability Model?

x P(x)

1 .2

2 -.3

3 .6

4 .1

x P(x)

A .3

B .2

C .5

D 0

x P(x)

Red .2

Green .2

Blue .3

Yellow .2

No, cannot have negative

probabilities

Yes No, the sum of the probabilities

doesn’t equal 1 5.1

A survey of 971 investors asked how often they tracked their portfolio. The table shows the investor responses. What is the probability that an investor tracks his or her portfolio daily?

How Frequently Response

Daily 236

Weekly 261

Monthly 273

Couple times a year 141

Don’t Track 60

Step 1: Find the total of responses = 971 Step 2: Take the amount of daily responses and divide by the total 236/971=0.243

Probability

5.1

Addition Rules and Disjoint Events

• What are disjoint events?

– AKA mutually exclusive

– Events that have no outcomes in common

• Addition Rule for Disjoint Events

– If E and F are disjoint events, then

P(E or F)=P(E)+ P(F)

• General Addition Rule

– P(E or F)=P(E)+ P(F)- P(E and F)

5.2

• In the game of craps, two dice are tossed and the up faces are totaled. Is the event getting a total of 9 and one of the dice showing a 6 disjoint events? Answer Yes or No. – No because you can get a sum of 9 when of the dice is

showing a 6 (the other would be a 3)

• A card is drawn from a standard deck of 52 playing cards. Find the probability that the card is a queen or a club. Express the probability as a simplified fraction. P(Queen or club)=P(Queen)+P(club)-P(Queen and a club)

P(Queen or club)= 4/52 + 13/52 – 1/52

P(Queen or club)= 16/52=4/13

Addition Rules and Disjoint Events

5.2

Compliment Rule P(not E)=1-P(E)

or

P(EC)=1-P(E)

5.2

Rooms Probability

1 .005

2 .011

3 .088

4 .183

5 .23

6 .204

7 .123

8 or More .156

• What Is the probability of having less than 8 rooms?

𝑃 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 8 = 1 − 𝑃 8 𝑜𝑟 𝑚𝑜𝑟𝑒= 1 − .156 = .844

• What is the probability of having at least 3 rooms?

𝑃 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 3= 1 − 𝑃 1 𝑟𝑜𝑜𝑚 − 𝑃 2 𝑟𝑜𝑜𝑚𝑠= 1 − .005 − .011 = .984

Independence and Multiplication Rule

• When are 2 events independent?

– If the occurrence of event E in a probability experiment does not affect the probability of event F

• Multiplication Rule for independent Events

– Events E and F are independent if

P(E and F) =P(E)*P(F)

5.3

Independence and Multiplication Rule

There are 30 chocolates in a box, all identically shaped. There are 11 filled with nuts, 10 filled with caramel, and 9 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Is this an example of independence? Answer Yes or No.

– No

A single die is rolled twice. Find the probability of getting a 2 the first time and a 2 the second time. Express the probability as a simplified fraction.

– These events are independent P(2 on the first and 2 on second)=P(2 on first roll)(2 on second roll)=(1/6)*(1/6)=1/16

5.3

Conditional Probability

• P(F|E) is read probability of event F given that event E has occurred.

• Conditional Probability Rule

– P F E =P(E and F)

P(E)=

N(E and F)

N(E)

• General Multiplication Rule

– P(E and F)=P(E)*P(F|E)

5.4

Conditional Probability Using the given table, given that the car selected was a domestic car, what is the probability that it was older than 2 years?

Age of car (in years)

Make 0-2 3-5 6-10 Over 10 Total

Foreign 40 30 10 20 100

Domestic 45 27 11 17 100

Total 85 57 21 37 200

P older than 2 𝑦𝑒𝑎𝑟𝑠 𝐷𝑜𝑚𝑒𝑠𝑡𝑖𝑐 =N(older than 2 years and Domestic)

N(Domestic)

=27 + 11 + 17

100=

11

20

5.4

General Multiplication Rule

There are 36 chocolates in a box, all identically shaped. There 10 are filled with nuts, 12 with caramel, and 14 are solid chocolate. You randomly select one piece, eat it, and then select a second piece. Find the probability of selecting solid chocolate then a nut.

P(solid and nut)=P(solid)*P(nut|solid)= 14

36∗

10

35=

1

9

5.4

Counting Techniques: Combination The number of different arrangements of r objects chosen from in which

1. The n objects are distinct

2. Repetition of objects is not allowed

3. Order is not important

Formula: nCr=𝑛!

𝑟! 𝑛−𝑟 !

From 9 names on a ballot, a committee of 3 will be elected to attend a political national convention. How many different committees are possible?

n=9, r=3

9C3=9!

3! 9−3 != 84

5.5

Counting Techniques: Permutations

• Permutation of Distinct items with replacement – The selection of r objects from n distinct objects, repetition is

allowed, and order matters – nr

• Permutation of Distinct items without replacement – The selection of r objects from n distinct objects, no repetition,

and order matters

– nPr=𝑛!

𝑛−𝑟 !

• Permutation of nondistinct items without replacement – The number of ways n objects can be arranged (order matters)

in which there are n1 of one kind, n2 of a second kind and so on where n=n1+n2+…+nk

–𝑛!

𝑛1!𝑛2!…𝑛𝑘!

5.5

Counting Techniques • Outside a home there is a keypad that will open the garage if the

correct 4-digit code is entered. How many codes are possible? – Permutation of Distinct items with replacement; 104= 10,000

• Suppose 40 cars start at the Indy 500. In how many ways can the top 3 cars finish? – Permutation of Distinct items without replacement; 40P3= 59280

• A family has 6 children. If this family has exactly 2 boys, how many different birth/gender orders are possible? – Combination; 6C2= 15

• A baseball team consists of 3 outfielders, 4 infielders, a pitcher, and a catcher. Assuming that the outfielders and infielders are indistinguishable, how many batting orders are possible?

– Permutation of no distinct items without replacement; 9!

3!4!1!1!= 2520

5.5

Determine the Appropriate Counting Technique to Use

5.5 Copyright © 2013 Pearson Higher Ed

5-17

Determine the Appropriate Probability to Rule to Use

Copyright © 2013 Pearson Higher Ed 5.6

Mean and Standard Deviation of Discrete Probability Model

x P(x)

0 .3

1 .4

2 .2

3 .06

4 .04

In a sandwich shop, the following probability distribution was obtained. The random variable x represents the number of condiments used for a hamburger. Find the mean and standard deviation for the random variable x.

Expected Value = Mean 1) Multiply x by P(x) 2) Add the numbers from step 1 0+.4+.4+.18+.16=1.14. This is the mean. Standard Deviation 1) Find the mean and square it, =1.14 2=1.2996

2) Take x2 and multiply by P(x) 3) Add the values from step 2 = 2.38 4) Take the number from step 3 and subtract from value from step 1: 2.38-1.2996=1.0804 5) Take the square root of the number from step 4, 1.04. This is the standard deviation.

xP(x)

0*.3=0

.4

.4

.18

.16

x2P(x)

02*.3=0

.4

.8

.54

.64

6.1

Binomial Probability Criteria for Binomial

1) Fixed number of trials (times done) 2) Trials are independent 3) Outcomes: Success or Failure 4) The probability of a success is the same

Decide whether the experiment is a binomial experiment. If it is not, explain why. 1) You draw a marble 350 times from a bag with three colors of marbles. The random variable represents the color of marble that is drawn. No, more than two outcomes 2) Selecting five cards, one at a time without replacement, from a standard deck of cards. The random variable is the number of picture cards obtained. No, Trials are not independent 3) Survey 50 college students see whether they are enrolled as a new student. The random variable represents the number of students enrolled as new students. Yes

6.2

Binomial Probability

Assume that male and female births are equally likely and that the births are

independent.

Find the probability that there are exactly 9 girls out of 10 births

𝑃 𝑋 = 9 = 10𝐶9 .5 9(1 − .5)10−9 = .0098

Find the probability that there is at least 1 girl out of 10 births

𝑃 𝑥 ≥ 1 = 1 − 𝑃 0 = .999

6.2

Binomial Mean and Standard Deviation

In 2000, it was reported that 11% of adult Americans had trust in the federal government’s handling of domestic issues. Suppose a random sample of 1100 finds that 84 have a trust in the government’s handling of domestic issues. Would these results be considered unusual? 1) Find the Mean: (np) =1100*.11=121

2) Find the standard deviation: 𝑛𝑝(1 − 𝑝) = 1100(.11)(1 − .11) =10.38

3) For an event to be unusual, it must be outside the mean plus/minus two standard deviations

121-2(10.28)=100.44; 121+2(10.28)=141.56 Since 84 is outside this interval, the event is unusual.

6.2

Poisson Distribution

• A stats professor finds that when he schedules an office hour at the 10:30 a.m. time slot, an average of three students arrive. Use the Poisson distribution to find the probability that in a randomly selected office hour in the 10:30 a.m. time slot exactly seven students will arrive. – x=7, λ=3, t=1

– 𝑃 𝑥 = 7 =(3∙1)7

7!𝑒−3∙1 =.0216

6.3