B Heard
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 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
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Let’s look at some questions….
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 How many ways can a committee of 4 be chosen from
20 people?
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 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.
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 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)?
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 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.
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 What values can a probability be?
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 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.
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 Given the following table of College Students by Gender
and Major:
 Answer the following questions.
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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?
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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
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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?
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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.
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 List the sample space of the National League teams in
the 2008 MLB playoffs.
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List the sample space of the National League teams in the
2008 MLB playoffs.
{Brewers, Cubs, Dodgers, Phillies}
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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?
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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
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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?
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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
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 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?
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 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
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 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.
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 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.
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Factorials
Answer the following:
4!
3! * 0!
2! /0!
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Factorials
Answer the following:
Remember that the factorial sign means x! = x * x-1 * x-2
* … 1, so
4! = 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)
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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.
Not to be used, posted, etc. without my expressed permission. B Heard
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).
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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?
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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.
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Find P(X < 14) for this random variable.
X = {1, 5, 7, 13, 15}.
P(1) = P(5) = P(7) = P(13) = P(15).
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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
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 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?
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 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.
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 We have a binomial experiment with p = .6 and n =
3. Set up the probability distribution and compute the
mean, variance, and standard deviation.
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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) = .22
E(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)
Not to be used, posted, etc. without my expressed permission. B Heard
Not to be used, posted, etc. without my expressed permission. B Heard
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)
Not to be used, posted, etc. without my expressed permission. B Heard
 We have the random variable X = {5,10} with P(5) = .6
and P(10) = .4. Find E(X).
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 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
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Continuous or discrete?
1.
2.
3.
4.
5.
6.
The amount of oil in your car’s engine?
The number of cans of coke in your refrigerator?
Your son’s weight?
The number of cousins you have?
The amount of butter in your butter dish?
The number of classes you have taken and received
credit for?
Not to be used, posted, etc. without my expressed permission. B Heard
Continuous or discrete?
1. The amount of oil in your car’s engine? Continuous
2. The number of cans of coke in your refrigerator?
Discrete
3. Your son’s weight? Continuous
4. The number of cousins you have? Discrete
5. The amount of butter in your butter dish?
Continuous
6. The number of classes you have taken and received
credit for? Discrete
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S
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