Forum
M6 L1
First Post
M6 L2
Look at the problem scenario given
below and answer the questions
given, then explain how you
obtained your answers. Post your
answers to the Forum. After you
make your post you'll be able to see
others' answers.
You can post your answers by
clicking on "Reply" below. After
you make your post you'll be able
to see others' answers. Check back
periodically to see who else has
posted. Make comments on the
correctness or errors to answers
given by two other classmates.
Problem Scenario:
Use the statement below to answer
the three questions given. It may be
useful to draw a Venn Diagram on
your own paper to work out the
problem.
First reply
In your post, answer the following
questions. After you make your post
you'll be able to see others' answers.
After you have filled in the wiki from
the first part of this assignment,
answer questions 3 and 4 from
Activity 6 on pp. 328-329 (TPOS 2nd
Edition) or Activity 6C pp. 406-407
(TPOS 3rd Edition) in your textbook.
Please submit your answers to the
forum and include how many trials
combined (yours and other students)
you used to get your answers.
1.
How many of the Chicagoans surveyed didn’t like any of the three teams? Justify your
answer with statistical explanations and calculations.
115 - 82 = 33. So, 33 Chicagoans did not like any of the three teams. (The 82 comes from
adding all the parts of the Venn Diagram together)
2. How many of the Chicagoans surveyed liked both the Cubs and Bulls? Justify your answer
with statistical explanations and calculations.
4
3. If one of these Chicagoans likes the Cubs, what's the probability he also likes the White
Sox? Justify your answer with statistical explanations and calculations.
P(liking white sox | likes Cubs) = P(liking cubs and white sox) / P(liking cubs)
= 12 / 35 = 0.343
In a recent survey of 115
Chicagoans, the following
information was obtained:
35 liked the Cubs
12 Liked both the Cubs and White
Sox
30 liked the Bulls
1 liked all three
55 liked the White Sox
23 liked both the Bulls and White Sox
4 liked only the Bulls
1.
2.
3.
M6 L3
How many of the
Chicagoans surveyed didn’t
like any of the three teams?
Justify your answer with
statistical explanations and
calculations.
How many of the
Chicagoans surveyed liked
both the Cubs and Bulls?
Justify your answer with
statistical explanations and
calculations.
If one of these Chicagoans
likes the Cubs, what's the
probability he also likes the
White Sox? Justify your
answer with statistical
explanations and
calculations.
Look at the problem scenario given
below and answer the questions
given, then explain how you
obtained your answers. Post your
answers to the Forum. After you
True or False
1. A simple event has only one possible outcome, whereas a compound event has multiple
outcomes. True
2. The sample space contains all the possible outcomes of a compound event. True
3. The closer to 1 the value of a probability is, the less likely it is that the event will occur.
make your post you'll be able to see
others' answers.
You can post your answers by
clicking on "Reply" below. After
you make your post you'll be able
to see others' answers. Check back
periodically to see who else has
posted. Make comments on the
correctness or errors to answers
given by two other classmates.
Problem Scenario:
True or False
1. A simple event has only one
possible outcome, whereas a
compound event has multiple
outcomes. ________
2. The sample space contains all the
possible outcomes of a compound
event. ________
3. The closer to 1 the value of a
probability is, the less likely it is that
the event will occur. ________
4. Simple events are always mutually
exclusive. ________
5. Adding up the probabilities of all
the simple events in an experiment
gives us 1. ________
6. Compound events are always
mutually exclusive. ________
7. The intersection of two events
contains all the outcomes contained in
the two events combined. ________
8. Events that share an intersection
are not mutually exclusive. ________
9. The probability of the union of two
events is equal to the sum of the
False The closer to 1 the value of a probability is, the more likely it is that the event will occur.
4. Simple events are always mutually exclusive. True
5. Adding up the probabilities of all the simple events in an experiment gives us 1. True
6. Compound events are always mutually exclusive. False Compound events can be either
mutually exclusive or not mutually exclusive.
7. The intersection of two events contains all the outcomes contained in the two events combined.
False
The intersection of two events contains all the outcomes contained in BOTH events.
8. Events that share an intersection are not mutually exclusive. True
9. The probability of the union of two events is equal to the sum of the probabilities of each
of the two events. False The probability of the union of two events is equal to the sum of the
probabilities of each of the two events, minus the probability of the intersection of the two events.
10. The complement of an event A is the event that A occurs. False The complement of an event
A is the event that A does NOT occur.
Problem Solving
11. P(less than 3) = P(rolling a 1) + P(rolling a 2) = 1/6 + 1/6 = 2/6 = 1/3
12. We have:
P(A) = P(rolling a 1) + P(rolling a3) +P(rolling a 5) = 1/6 +1/6 +1/6 = 3/6 = 1/2
P(B) = 1/3 (calculated above)
P(AB) = P(rolling a 1) = 1/6
So P(A∪B) = P(A) +P(B) – P(AB) = ½ + 1/3 – 1/6 = 5/6 – 1/6 = 4/6 = 2/3
13. We have that P(A) + P(complement of A) = 1, so P(A) = 1 – P(complement of A) = 1 - .3 = .6
14. P(A|B) = P(AB)/P(B) = (1/6)/(1/3) = (1/6).(3/1) = ½
2.
probabilities of each of the two
events. ________
10. The complement of an event A is
the event that A occurs. _________
Problem Solving
11. Find the probability of getting a
number less than 3 when rolling a
single six-sided fair die.
12. Find the probability of the union
and the probability of the intersection
of the following events:
 A = getting an odd number
when rolling a die
 B = getting a number less
than 3
13. If the probability of the
complement of A is 0.3, what is the
probability of A occurring?
14. Find the conditional probability of
A given B, using events A and B from
12 above.
M7 L1
random
variables
In your post, answer
the following
questions. After you
make your post you'll
be able to see others'
answers.
TPOS 2nd edition:
TPOS 2nd Edition
7.1) a) P(less than 3) = P(1 or 2) = (2/6) = (1/3).
b-c) answers will vary
7.2(a)
BBB BGB GBB GGB
BBG BGG GBG GGG
1/8 probability each
(b) 3/8
Read pp. 487-502. Work
problems 9.1, 9.2, 9.3, 9.4,
9.5.
TPOS 3rd edition:
Read pp. 563-579. Work
problems 9.1, 9.2, 9.3
(c)
X
P(X)
0
1/8
1
3/8
2
3/8
3
1/8
7.3) see back of book for answers.
7.6) (a) P(0 ≤ X ≤ 0.4) = 0.4
(b) P(0.4 ≤X ≤ 1) = 0.6
(c) P(0.3 ≤ X ≤ 0.5) = 0.2
(d) P(0.3 < X < 0.5) = 0.2
(e) P(0.226 ≤ X ≤ 0.713) = 0.487
(f) A continuous distribution assigns probability 0 to every individual
outcome. In this case, the probabilities in (c) and (d) are the same because
the events differ by 2 individual values, 0.3 and 0.5, each of which has
probability 0.
TPOS 3rd Edition:
7.1) a) P(less than 3) = P(1 or 2) = (2/6) = (1/3).
b-c) answers will vary
7.2) (a)
BBB BGB GBB GGB
BBG BGG GBG GGG
1/8 probability each
(b) 3/8
(c)
X
0
1
2
3
P(X)
1/8
3/8
3/8
1/8
7.3) a) (X ≥ 1) or (X > 0). P(X ≥ 1) = 1 – P(X<1) = 1-P(X = 0) = 1 – 0.1 = 0.9
b) The event (X ≤ 2) is “no more than two nonword errors,” or “fewer than
three nonword errors.” P(X ≤ 2) = 0.6. P(X < 2) = 0.3.
7.8) (a) P(0 ≤ X ≤ 0.4) = 0.4
(b) P(0.4 ≤X ≤ 1) = 0.6
(c) P(0.3 ≤ X ≤ 0.5) = 0.2
(d) P(0.3 < X < 0.5) = 0.2
(e) P(0.226 ≤ X ≤ 0.713) = 0.487
(f) A continuous distribution assigns probability 0 to every individual
outcome. In this case, the probabilities in (c) and (d) are the same because
the events differ by 2 individual values, 0.3 and 0.5, each of which has
probability 0.
M7 L1 law of
large
numbers
In your post, answer the following
questions. After you make your post
you'll be able to see others' answers.
TPOS 2nd Ed:
Work problems 7.30, 7.32, 7.34.
TPOS 3rd Ed:
Work problems 7.31, 7.32, 7.38
TPOS 2nd Edition
7.30 The graph for xmax = 10 displays visible variation for the first ten values
of x, whereas the graph for xmax = 100 gets closer and closer to
=64.5 as x increases. This illustrates that the larger the sample size
(represented by the integers 1,2,3,?in L1), the closer the sample mean gets
to population mean. This illustrates the Law of Large Numbers.
7.32(a) The wheel is not affected by previous outcomes - it has no memory;
outcomes are independent. So on any one spin, black and red remain
equally likely.
(b) Wrong. Removing a card changes the composition of the remaining
deck, so successive draws are not independent. If you hold 5 red cards, the
deck now contains 5 fewer red cards, so your chance of another red
decreases.
7.34(a) Independent: Weather conditions a year apart should be
independent.
(b) Not independent: Weather patterns tend to persist for several days;
today's weather tells us something about tomorrow's.
(c) Not independent: The two locations are very close together, and would
likely have similar weather conditions.
TPOS 3rd Edition
7.31)The graph for xmax = 10 displays visible variation for the first ten values
of x, whereas the graph for xmax = 100 gets closer and closer to
=64.5 as x increases. This illustrates that the larger the sample size
(represented by the integers 1,2,3,?in L1), the closer the sample mean gets
to population mean. This illustrates the Law of Large Numbers.
7.32) (a) The wheel is not affected by previous outcomes - it has no
memory; outcomes are independent. So on any one spin, black and red
remain equally likely.
(b) Wrong. Removing a card changes the composition of the remaining
deck, so successive draws are not independent. If you hold 5 red cards, the
deck now contains 5 fewer red cards, so your chance of another red
decreases.
7.38) (a) Independent: Weather conditions a year apart should be
independent.
(b) Not independent: Weather patterns tend to persist for several days;
today's weather tells us something about tomorrow's.
(c) Not independent: The two locations are very close together, and would
likely have similar weather conditions.
Graphing
Calculator
activity and
discussion
Discussion
assessment
In your post, answer the
following questions. After
you make your post you'll
be able to see others'
answers.
TPOS 2nd Ed:
Problems 7.1 and 7.6
TPOS 2nd Edition
7.1) a) P(less than 3) = P(1 or 2) = (2/6) = (1/3).
b-c) answers will vary
7.2(a)
BBB BGB GBB GGB
BBG BGG GBG GGG
1/8 probability each
(b) 3/8
TPOS 3rd Ed:
Problems 7.1 and 7.8
(c)
X
P(X)
0
1/8
1
3/8
2
3/8
3
1/8
7.3) see back of book for answers.
7.6) (a) P(0 ≤ X ≤ 0.4) = 0.4
(b) P(0.4 ≤X ≤ 1) = 0.6
(c) P(0.3 ≤ X ≤ 0.5) = 0.2
(d) P(0.3 < X < 0.5) = 0.2
(e) P(0.226 ≤ X ≤ 0.713) = 0.487
(f) A continuous distribution assigns probability 0 to every individual
outcome. In this case, the probabilities in (c) and (d) are the same
because the events differ by 2 individual values, 0.3 and 0.5, each of
which has probability 0.
TPOS 3rd Edition:
7.1) a) P(less than 3) = P(1 or 2) = (2/6) = (1/3).
b-c) answers will vary
7.2) (a)
BBB BGB GBB GGB
BBG BGG GBG GGG
1/8 probability each
(b) 3/8
(c)
X
0
1
2
3
P(X)
1/8
3/8
3/8
1/8
7.3) a) (X ≥ 1) or (X > 0). P(X ≥ 1) = 1 – P(X<1) = 1-P(X = 0) = 1 – 0.1 = 0.9
b) The event (X ≤ 2) is “no more than two nonword errors,” or “fewer than
three nonword errors.” P(X ≤ 2) = 0.6. P(X < 2) = 0.3.
7.8) (a) P(0 ≤ X ≤ 0.4) = 0.4
(b) P(0.4 ≤X ≤ 1) = 0.6
(c) P(0.3 ≤ X ≤ 0.5) = 0.2
(d) P(0.3 < X < 0.5) = 0.2
(e) P(0.226 ≤ X ≤ 0.713) = 0.487
(f) A continuous distribution assigns probability 0 to every individual
outcome. In this case, the probabilities in (c) and (d) are the same
because the events differ by 2 individual values, 0.3 and 0.5, each of
which has probability 0
Graphing
Calculator
activity and
discussion
In this activity, you did 5-10 trials of
200 with your lists. In your post,
(1) Give the area of the circle that you
found for each trial. So you will be
listing 5-10 areas.
(2) Compare your areas with other
students' areas. You may have to
come back later to do this. Using the
math you know from before this
course, what should the exact area of
the circle be?
(1) Answers for each trial will vary.
(2) The area of the circle should be equal to π, since the radius of the circle is 1 unit, and the formula
for the area of a circle is πr2
M7 L1
random
Variables
In your post, answer the following
questions. After you make your post
you'll be able to see others' answers.
link pdf http://152.46.13.240/MoodleContent/APstats/M7%20L1%20HW%20Answers.pdf
M7 L2 Rules
for means
In your post, answer the following
questions. After you make your post
Compare
your results
TPOS 2nd Ed: Problems
7.8 and 7.24
TPOS 3rd Ed: Problems 7.9
and 7.26
TPOS 2nd Edition
7.35) a) Dependent: since the cards are being drawn from the deck without
and
variances
you'll be able to see others' answers.
TPOS 2nd Ed:
Work problems 7.35, 7.36, 7.37,
TPOS 3rd Ed:
Work problems 7.39, 7.40, 7.41,
replacement, the nature of the third card (and thus the value of Y) will
depend upon the nature of the first two cards that were drawn (which
determine the value of X)
b) Independent: X relayes to the outcome of the first roll, Y to the outcome
of the second roll, and individual dice rolls are independent (the dice have
no memory).
7.36. The total mean is 40+5+25 = 70 min.
7.38. Assuming that the two times are independent, the total variance
is
, so
. Assuming that
the two times are dependent with correlation .3, the total variance
is
,
so
. The positive correlation of .3 indicates that the two
times have some tendency to increase together or decrease together, which
increases the variability of their sum.
TPOS 3rd Edition
7.39) a) Dependent: since the cards are being drawn from the deck without
replacement, the nature of the third card (and thus the value of Y) will
depend upon the nature of the first two cards that were drawn (which
determine the value of X)
b) Independent: X relates to the outcome of the first roll, Y to the outcome of
the second roll, and individual dice rolls are independent (the dice have no
memory).
7.40) The total mean is 40 + 5 + 25 = 70 minutes.
7.41) see back of book.
M7 L2 Means In your post, answer the following
questions. After you make your post
for random
you'll be able to see others' answers.
TPOS 2nd Edition
7.55) The missing probability is 0.99058 (so that the sum is 1). This gives
variables
TPOS 2nd Ed:
Work problems 7.55, 7.56.
TPOS 3rd Ed:
Work problems 7.55, 7.56.
mean earnings of $303.3525.
7.56. The mean μ of the companies "winnings" (premiums) and their
"losses" (insurance claims) is positive. Even though the company will lose a
large amount of money on a small number of policyholders who die, it will
gain a small amount on the majority. The law of large numbers says that the
average "winnings" minus "losses" should be close to μ, and overall the
company will almost certainly show a profit. In this case it will gain an
average of $303.35 per insurance policy.
TPOS 3rd Edition
7.55) The missing probability is 0.99058 (so that the sum is 1). This gives
mean earnings of $303.3525.
7.56. The mean μ of the companies "winnings" (premiums) and their
"losses" (insurance claims) is positive. Even though the company will lose a
large amount of money on a small number of policyholders who die, it will
gain a small amount on the majority. The law of large numbers says that the
average "winnings" minus "losses" should be close to μ, and overall the
company will almost certainly show a profit. In this case it will gain an
average of $303.35 per insurance policy.
M7 Review
Read the material below then answer
the questions in your post. After you
post, you will be able to see others'
answers.
TPOS 2nd Ed:
Read pp. 430-431. Work
TPOS 2nd edition AND TPOS 3rd Edition
7.57) The variance is 94,236,826.64 so the standard deviation is the square
root of that = $9797.57
problems 7.57, 7.58, 7.61
TPOS 3rd Ed:
Read pp. 504-505. Work
problems 7.57, 7.58, 7.61
7.58(a)
(b) With this new definition of Z:
(unchanged),
,
(smaller by a factor of
)
7.61) see back of book