Week 5 Lecture Statistics For Decision Making

B Heard

Not to be used, posted, etc. without my expressed permission. B Heard

Your Week 5 Quiz is on material covered in Weeks 3 and 4

Your Week 7 Quiz is on material covered in Weeks 5 and 6

Your Final Exam is comprehensive covering the material in the three prior quizzes plus the material covered in Week 7

Your best approach for preparing for the quizzes should be the Practice Quizzes offered in the previous week (for this week’s quiz the Week 2 Practice Quiz) and the live lecture of the Week the actual Quiz is posted


Let’s look at some questions….


How many ways can a committee of 4 be chosen from 20 people?


How many ways can a committee of 4 be chosen from 20 people?

This would be a combination because “order” doesn’t matter, so there would be 4845 different ways.


How many ways can a committee of 4 be chosen from 20 people if they have distinct positions (i.e. President, Secretary, Treasurer, and Vice-President)?


How many ways can a committee of 4 be chosen from 20 people if they have distinct positions (i.e. President, Secretary, Treasurer, and Vice-President)?

This would be a permutation because “order” does matter, so there would be 116280.


What values can a probability be?


What values can a probability be?

Anything between 0 and +1 (NOTHING ELSE). That also means from 0% to 100%, and any positive fraction where the numerator is smaller than the denominator.


Given the following table of College Students by Gender and Major:

Answer the following questions.



1. What is the probability that a given student is a female?2. What is the probability that a given student is a Math Major?3. What is the probability that a given student is male Political

Science Major? (This could also be worded “What is the probability that a student is a Political Science Major AND Male)

4. What is the probability that a student is a Business Major given that the student is female?


1. What is the probability that a given student is a female? 76/150 or .5067

2. What is the probability that a given student is a Math Major? 10/150 or .0667

3. What is the probability that a given student is male Political Science Major? (This could also be worded “What is the probability that a student is a Political Science Major AND Male) 20/150 or .1333

4. What is the probability that a student is a Business Major given that the student is female? 22/76 or .2895


1. What is the probability that a student is a Male given that the student is an English Major?

2. What is the probability that a student is female or a Psychology Major?

3. What is the probability that a student is a Business Major or a male?

4. Are being male and being a Business Major independent events?


1. What is the probability that a student is a Male given that the student is an English Major? 12/25 or .4800

2. What is the probability that a student is female or a Psychology Major? (76 + 35 – 19)/150 = 92/150 or .6133

3. What is the probability that a student is a Business Major or a male? (45 + 74 – 23)/150 = 96/150 or .64

4. Are being male and being a Business Major independent events? If P(male| Business Major) = P(Male) they are independent events. However the probability of being a male student given that they are a business major is 23/25 and the probability of being a male student is 74/150, they are not equal thus these are DEPENDENT events.

List the sample space of the National League teams in the 2008 MLB playoffs.



List the sample space of the National League teams in the 2008 MLB playoffs.

{Brewers, Cubs, Dodgers, Phillies}


What is the probability of drawing a 7 from a deck of cards? And what is the probability of a second card being an Ace or King if the first was a 7?


What is the probability of drawing a 7 from a deck of cards? And what is the probability of a second card being an Ace or King if the first was a 7?

What is the probability of drawing a 7 from a deck of cards? 4/52 or 1/13 And what is the probability of a second card being an Ace or King if the first was a 7?There are 8 Aces and Kings left, but only 51 cards to draw from so it would be 8/51


What is the probability of drawing a 6, 7, or 8 from a deck of cards? What is the probability of a second card drawn being a 6, 7, or 8 if the first was a 6, 8, or 8?


What is the probability of drawing a 6, 7, or 8 from a deck of cards? What is the probability of a second card drawn being a 6, 7, or 8 if the first was a 6, 8, or 8?

What is the probability of drawing a 6, 7, or 8 from a deck of cards? There would be 12 of them so 12/52 or 3/13 What is the probability of a second card drawn being a 6, 7, or 8 if the first was a 6, 8, or 8?There would be 11 left and only 51 cards to draw from so it would be 11/51

If there are 13 word documents and 27 excel documents in a folder, and one is randomly drawn, what is the probability of drawing a word document?


If there are 13 word documents and 27 excel documents in a folder, and one is randomly drawn, what is the probability of drawing a word document?

13/ (13+17) = 13/40


If the P(A) = 7/20 ; P(B) = 14/20; P(A and B) = 3/20, then what is the P(A OR B)? Are A and B mutually exclusive? Prove your answer.


If the P(A) = 7/20 ; P(B) = 14/20; P(A and B) = 3/20, then what is the P(A OR B)? Are A and B mutually exclusive? Prove your answer.

P(A or B) = P(A) + P(B) – P(A and B) = 7/20 + 14/20 – 3/20 = 18/20 or 9/10

A and B are NOT mutually exclusive because P(A and B) does not equal zero.


FactorialsAnswer the following: 4! 3! * 0! 2! /0!


FactorialsAnswer the following: Remember that the factorial sign means x! = x *

x-1 * x-2 * … 1, so4! = 4*3*2*1 = 24 3! * 0! = (3*2*1) * 1 = 6 (remember 0! is ALWAYS

= 1) 2! /0! = (2*1)/1 = 2 (remember 0! is ALWAYS =

1)Not to be used, posted, etc. without my expressed permission. B Heard


Decide whether the following experiments would be Binomials, Poissons, or neither. 1.You test 6 different types of batteries. The random variable represents the battery that is last longest. Past experience is that 30% of the time it is the third of the six types. 2.You observe a stop sign for 4 hours. The random variable represents the number of cars that either completely stopped or didn’t. Historically 65% of cars come to a complete stop. 3. A cab company averages three pickups per hour. We're interested in knowing the probability that in a randomly selected hour they will get one pickup.4. A company ships computer components in boxes that contain 20 items. We want to know the probability that the 2nd item removed will be defective.


1.You test 6 different types of batteries. The random variable represents the battery that is last longest. Past experience is that 30% of the time it is the third of the six types. Neither, because we are testing 6 different types (it’s not a yes/no, good/bad, two decision type situation)

2.You observe a stop sign for 4 hours. The random variable represents the number of cars that either completely stopped or didn’t. Historically 65% of cars come to a complete stop. Binomial, probability given in percentage. For this to be Poisson it would say something like on average 42 cars stop at the stop sign every hour, we want to know the probability of exactly 32 stopping, or more than 45 stopping, etc. – the probability (%) was a tip off that it was binomial

3. A cab company averages three pickups per hour. We're interested in knowing the probability that in a randomly selected hour they will get one pickup. Poisson, as per the previous question’s answer we are interested in finding out the probability of 1 pickup.

4. A company ships computer components in boxes that contain 20 items. We want to know the probability that the 2nd item removed will be defective. Neither, we don’t have a probability to start with (Binomial), or an average number of defects (Poisson).


If X = {1, 5, 9, 12} and P(1) = .3, P(5) = .3, P(9) = .2, and P(12) = .2, can we call it a random variable?


If X = {1, 5, 9, 12} and P(1) = .3, P(5) = .3, P(9) = .2, and P(12) = .2, can we call it a random variable?

Yes, the sum of the probabilities = (.3+.3+.2+.2) = 1 and they are all between 0 and 1.

Find P(X < 14) for this random variable. X = {1, 5, 7, 13, 15}. P(1) = P(5) = P(7) = P(13) = P(15).


Find P(X < 14) for this random variable. X = {1, 5, 7, 13, 15}. P(1) = P(5) = P(7) = P(13) = P(15).

Since P(1) = P(5) = P(7) = P(11) = P(13) = P(15) they must add up to 1 therefore the probability for each must be 1/5 since there are five so it is 0.20

then P(x < 14) = P(1) + P(5) + P(7) + P(13) = 0.20 + 0.20 + 0.20 + 0.20 = 0.80


If X = {-1, 0, 3, 8} and P(-1) = .3, P(0) = .1, P(3) = .3, and P(8) = .3, can we call it a random variable?


If X = {-1, 0, 3, 8} and P(-1) = .3, P(0) = .1, P(3) = .3, and P(8) = .3, can we call it a random variable?

Do the probabilities add up to one? .3 + .1 + .3 +. 3 = 1 So yes it is (also note that those probabilities have to be between 0 and 1.


We have a binomial experiment with p = .6 and n = 3. Set up the probability distribution and compute the mean, variance, and standard deviation.


See Excel Spreadsheet picture that follows.

X = {0, 1, 2, 3}P(X = 0) = .0.06 P(X = 1) = .29 P(X = 2) = .43

P(X = 3) = .22E(X) = n*p = 3 * .6 = 1.8 (listed as mean in provided excel spreadsheet picture that follows) V(X) = n*p*q, q = 1 - p = 1 - .6 = .4 V(X) = 3*.6*.4 = .72 (listed as variance in provided excel spreadsheet)standard deviation = sqrt(variance) = sqrt(.72) = .85 (listed as stdev in provided excel spreadsheet picture that follows)



We have a Poisson with mu = 3. Find P(X = 4), find P(X < 4), find P(X <= 4), compute the mean, variance, and standard deviation.

See Excel Spreadsheet attached to follow on post.P(X = 4) = 0.168 (see picture of excel spreadsheet yellow block)P(X < 4) = 0.647 (see picture of excel spreadsheet green block)P(X <=4) = 0.353 (see picture of excel spreadsheet gray block)mean = variance = 3 (see picture of excel spreadsheet)standard deviation = sqrt(variance) = 1.73 (see picture of excel spreadsheet)


We have the random variable X = {5,10} with P(5) = .6 and P(10) = .4. Find E(X).


http://images.google.com/imgres?imgurl=http://focus.aps.org/files/focus/v19/st18/marbles_BIG.jpg&imgrefurl=http://focus.aps.org/story/v19/st18&h=482&w=724&sz=51&hl=en&start=9&um=1&usg=__I2i2WvGyphmck5c5bNo5ydMC54k=&tbnid=QMJjEiqbsfu5sM:&tbnh=93&tbnw=140&prev=/images%3Fq%3Dmarbles%26um%3D1%26hl%3Den

We have the random variable X = {5,10} with P(5) = .6 and P(10) = .4. Find E(X).

E(X) = sum of (x*P(X)) = 5*P(5) + 10*P(10) = 5*.6 + 10*.4 = 3.0 + 4.0 = 7.0


Continuous or discrete?

1.The amount of oil in your car’s engine?2.The number of cans of coke in your

refrigerator?3.Your son’s weight?4.The number of cousins you have?5.The amount of butter in your butter dish?6.The number of classes you have taken and

received credit for?


Continuous or discrete?1.The amount of oil in your car’s engine?

Continuous2.The number of cans of coke in your

refrigerator? Discrete3.Your son’s weight? Continuous4.The number of cousins you have? Discrete5.The amount of butter in your butter dish?

Continuous6.The number of classes you have taken and

received credit for? DiscreteNot to be used, posted, etc. without my expressed permission. B Heard

STAT CAVE

See you next week: “Same Stat Time, Same Stat Channel”Not to be used, posted, etc. without my expressed permission. B Heard

I will post charts at:

4stats.wordpress.com

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Week 5 Lecture Statistics For Decision Making