Total

Report
AP Statistics
February 27, 2012
AP Statistics B warm-up
Monday, February 27, 2012
1.
Write out the general addition rule of Chapter 15. How does it
differ from the addition rule of Chapter 14?
2.
30% of all Americans are overweight, 35% have high bloodpressure, and 23% are both overweight and suffer from high
blood pressure
a) What percentage of Americans have high blood pressure but
are not overweight?
b) What percentage of Americans either have high bloodpressure or are overweight, but are not suffering from both?
c) What percentage of Americans do not suffer from either high
blood-pressure or overweightness (which probably isn’t a word)?
Answers are on the next slide.
Slide 2 of 49
Answers to warmup
Monday, February 27, 2012
P( A  B)  P( A)  P( B)  P( A  B)
1.
.
2.
Draw a diagram (“Make a picture, make a picture, make a
picture!”) (C’mon, somebody go to the board and draw a Venn
diagram before I give you the answers!) Push the space bar to
display the answers.
a) 12% (which = 35% who have high blood-pressure minus
23% who are both overweight and suffer from high blood
pressure
b) 19% (Add the 12% from 2 a) to 7% who are overweight
but do not have high blood pressure (subtract 23% (both
overweight and high blood-pressure) from 30% who are
overweight))
c) 58% (42% have some combination of being overweight and
having high blood-pressure)
Slide 3 of 49
Outline for today
Monday, February 27, 2012
• Ms. Thien reports that you’d like to review problems 14, 15,
16, and 24 from Chapter 15. Most of today will be spent
going through these.
• I working at getting smaller sound files to embed in the
PowerPoints. Friday’s was 185 megabytes. I’m striving to
get them small enough (5-10 megs) so that I can post
them on the Garfield web site and you can download them
with the sound.
• Please text or email me comments….what you liked, what
you hated, just so that’s it’s feedback.
• You can stop the display, but make sure you go to the final
slide (#49) by the end of the period, since it has
tomorrow’s homework assignment.
Slide 4 of 49
Problem 14: birth order, take 2
Birth Order
First or only
Second or
later
Total
Arts and Sciences
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
Slide 5 of 49
How do we determine conditional
probabilities using a table?
(problem 14(b)
1. Pick the row (or column) that follows “given” or “for
A…..” or similar statement
E.g., here: “Among Arts and Sciences, students…..” (Ch.
15, problem 14(b))
2. Pick the column (or row) that corresponds to the
“probability”
Here, “the probability that a student was second child or
more”?
3. The intersection of the column and row in the table
is the numerator
4. The “total” at the end of the first column or row that
you picked in #1 above is the denominator
Slide 6 of 49
Yellow=first row we pick “Arts & Sciences”
Red=column we pick next
23=numerator
57=denominator
Birth Order
First or only
Second or
later
Total
Arts and Sciences
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
Answer: 23/57 = 40.35% or 0.4035
Slide 7 of 49
We could also do it by picking a column
first (yellow) and then a row (red)
• E.g., Given that a person is a second child, what
is the probability that they will be enrolled in
Agriculture? (this is similar to problem 14(c))
First or
only
Second
or later
Total
Arts and
Sciences
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
So here, the
probability is
41/110,
which equals
0.3727, or
37.27%.
Slide 8 of 49
Distinguishing probabilities
• What we’ve just done is a conditional
probability.
• The problems also ask for absolute
probabilities, i.e., the probabilities
that you will draw any specific person
at random.
• For this, the analysis is a bit
different.
Slide 9 of 49
Absolute (universal)
probabilities
• Somebody read problem 14(a) for
the class.
• Here, we need to establish the
probabilities for EVERYTHING inside
the table.
• The key is that we divide each entry
of the table by the total number of
students (that is, the entry in the
bottom row, last column)…here, 223.
Slide 10 of 49
So here’s what your table of
probabilities should look like
Birth Order
First or only
Second or
later
Total
Arts and Sciences
34 (34/223)
15.25%
23 (23/223)
10.31%
57
Agriculture
52 (52/223)
23.32%
41 (41/223)
18.39%
93
Human Ecology
15 (15/223)
6.73%
28 (28/223)
12.56%
43
Other
12 (12/223)
5.38%
18 (18/223)
30
Total
113
110
223
Note: it’s easier and faster to calculate this on Excel than with a
calculator!
Slide 11 of 49
Problems 14 (c), (d), and (e)
• (No audio on this slide)
• Take 5 minutes or so and recalculate
the probabilities for Problems (c), (d)
and (e)
• When you’re done, push the space
bar and I’ll show you another way of
calculating these.
Slide 12 of 49
Problem 14(c)
• Read problem, someone!
• Calculate probabilities of second children
• Pick A&S (yellow)
Birth Order
First or only
Second or later
Total
Arts and Sciences
34
23 (23/110)
20.9%
57
Agriculture
52
41 (41/110)
37.27%
93
Human Ecology
15
28 (28/110)
25.45%
43
Other
12
18 (18/110)
16.36%
30
Total
113
110
223
Slide 13 of 49
Problem 14(d)
• Pick 1st child column
• Calculate probability for Agriculture
First or only
Second or
later
Total
34
23
57
52 (52/113)
46.02%
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
Arts and Sciences
Agriculture
Slide 14 of 49
Problem 14: birth order, take 2
Birth Order
First or only
Second or
later
Total
Arts and Sciences
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
Slide 15 of 49
Problem 14(e)
What is the probability that an Agriculture student is a first or
only child? Answer: 52/93, or 55.91%
Birth Order
First or only
Second or
later
Total
Arts and Sciences
34
23
57
Agriculture
52
41
93
Human Ecology
15
28
43
Other
12
18
30
Total
113
110
223
Slide 16 of 49
Notation in terms of P(B|A)
Let’s rephrase problem 14 in terms of
conditional probability:
14(b): P(2nd child|Arts and Science)
14(c): P(Arts & Science|2nd child)
14(d): P(Agriculture|1st or only child)
14(e): P(1st or only child|Agriculture)
Slide 17 of 49
Comparison of P(B|A) with
language of problems (14(b))
P(B|A)
• 14(b): P(2nd child|Arts
and Science)
Problem
• Among the Arts and
Sciences students,
what’s the probabilibty
a student was a
second child (or
more)?
Comparison of P(B|A) with
language of problems (14(c))
P(B|A)
• 14(c): P(Arts &
Science|2nd child)
Problem
• Among second
children (or more),
what’s the probability
a student is enrolled
in the Arts and
Sciences?
Comparison of P(B|A) with
language of problems (14(d))
P(B|A)
• 14(d):
P(Agriculture|1st or
only child)
Problem
• What’s the probability
that a first or only
child is enrolled in the
Agriculture College?
Comparison of P(B|A) with
language of problems (14(e))
P(B|A)
• 14(e): P(1st or only
child|Agriculture)
Problem
• What is the probability
that an Agriculture
student is a first or
only child?
Problem 15
Sick Kids. Seventy percent of
kids who visit a doctor have a
fever, and 30% of kids with a
fever have sore throats. What’s
the probability that a kid who
goes to the doctor has a fever and
a sore throat?
Problem 15, sick kids: make a picture,
make a picture, make a picture!
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
Slide 23 of 49
Problem 15: sick kids
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
Slide 24 of 49
Problem 15: sick kids
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
30% of patients
who do not have
fever
This includes those
who have sore
throats but no
fever
Slide 25 of 49
Problem 15: sick kids
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
10%
10%
10%
10%
10%
10%
30% of patients
who do not have
fever
This includes those
who have sore
throats but no
fever
10%
10%
10%
10%
Slide 26 of 49
Problem 15: sick kids
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
30% of patients
who do not have
fever
This includes those
who have sore
throats but no
fever
Red=fever AND sore throat
Blue= fever alone
Slide 27 of 49
Problem 15 solution: just count the
boxes (each box = 1%) = 21%
70% of patients have fever; 30% of those with fever have
sore throat. % of both?
30% of patients
who do not have
fever
This includes those
who have sore
throats but no
fever
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Red=fever AND sore throat
Blue= fever alone
Slide 28 of 49
How can we solve this faster?
Answer: the General Multiplication
Rule, i.e.,
P( A  B)  P( A)  P( B | A)
Slide 29 of 49
General Rule of multiplication
P(AΩB) is what we are looking for, which=?
P(AΩB) = P(A) × P(B|A)
P(A)= probability of having a fever, which is?
70%
P(B|A) = probability of having a sore throat
GIVEN that you have a fever, which is?
30%
So the answer is 30% of 70%.....
………or 0.3 × 0.7 = 0.21 = 21%. QED.
Slide 30 of 49
Problem 16
Sick Cars. Twenty percent of cars
that are inspected have faulty
pollution control system. The
costs of repairing a pollution
control system exceeds $100
about 40% of the time. When a
driver takes a car in for
inspection, what’s the probability
that she will end up paying more
than $100 to repair the pollution
control system?
Let’s draw a diagram
(like problem 15)
Percentage of cars
Cars needing
repairs (20%)
Cars that need no repairs (80%)
Slide 32 of 49
Let’s draw a diagram
(like problem 15)
Need repairs
Don’t need repairs
>$100
<$100
Answer: 8 cells out of 100, or 8% (0.08)
Slide 33 of 49
Redoing problem 16 in
language/formulas
• The driver has a 20% chance of having
to repair the pollution control system
(P(A)).
• If she has to repair the pollution control
system, she has a 40% chance of
paying more than $100 (P(B|A))
• The general multiplication rule has us
multiply the two together: 0.2 × 0.4 =
0.08 = 8%.
• P(AΩB)=P(A) × P(B|A)
Slide 34 of 49
Problem 24
On the road again. According to Ex. 4,
the probability that a U.S. resident has
traveled to Canada is 0.18, to Mexico is
0.09, and to both countries is 0.04.
a) What is the probability that someone
who has traveled to Mexico has visited
Canada, too?
b) Are travel to Mexico and Canada disjoint
events? Explain.
c) Are travel to Mexico and Canada
independent events? Explain.
Draw the Venn diagram
0.77
Mexico
Canada
0.18-0.04=
0.14
0.04
0.09-0.04 =
0.05
Slide 36 of 49
Another approach
Travel to Mexico
Travel
to
Canada
Yes
No
Total
Yes
0.04
0.14
0.18
No
0.05
0.77
0.82
Total
0.09
0.91
1.00
Slide 37 of 49
Problem 24(a)
• What’s the probability that someone
who has travelled to Mexico has
travelled to Canada?
• Use either Venn diagram or table
– Total percent of people travelling to
Mexico? 9%
– Total percent of those travelling to
Mexico also travelling to Canada? 4%
• 4%/9%=0.444=44.4%
Slide 38 of 49
Problem 24(b)
• Are travel to Mexico and Canada
disjoint events? Explain.
• Disjoint=if A occurs, B cannot occur.
– Example: getting a grade. If you get an A,
you can’t get a B. You can have a
probability of getting an A, or a B, but you
can’t get both
• Travel is not disjoint because
– You can go to both
– 4% of the population HAS been to both
countries!
Slide 39 of 49
Problem 24(c)
• Are travel to Mexico and Canada
independent events? Explain.
• Independent=one event doesn’t alter
the probability of the other occurring
• Travel is not independent because
– 18% of all U.S. residents have been to
Mexico.
– 44.4% of those have also been to Canada
– If the events were independent, the
probabilities would be equal
Slide 40 of 49
Today’s lesson
• “Drawing without replacement”
• Usually done with decks of cards, but
lots of other uses as well
• Examples is on pp.354-55 of text
Slide 41 of 49
Visualizing the problem
• Draw a picture!
• 3 Gold (desirable), 4 Silver (less
desirable), 5 Wood (we don’t like)
Gold
Silver
Silver
Brown
Gold
Silver
Brown
Brown
Gold
Silver
Brown
Brown
Slide 42 of 49
You pick one and get gold
•
•
•
•
Your odds were 3/12 = ¼.
1 less gold (only 2 left)
1 less option (now only 11 choices)
What are the odds of your friend
selecting gold? Answer: 2/11
Silver
Silver
Brown
Gold
Silver
Brown
Brown
Gold
Silver
Brown
Brown
Slide 43 of 49
Multiplication rule applies
• Your odds of getting a Gold were ¼.
• The remaining odds of your friend
getting Gold on the next draw were
2/11.
• Multiply both together to get 2/44 or
1/22 = 0.045.
Slide 44 of 49
Problem 17 (p. 364)
• Rules: you draw three cards, one at
a time.
• We are looking at probabilities of
specific outcome.
• Trick: well, not a trick, exactly, but
another way of looking at things:
recast the question not as the odds
of GETTING something, but of NOT
getting it.
Slide 45 of 49
Example: 17(a)
(first heart you get is the third card dealt)
• The odds of getting a heart are ¼ for
the first card dealt (13 hearts out of
52 cards in all)
• But the probability is that you
DIDN’T get a heart. That’s ¾, or .75
(because there are 39 cards in the
deck that aren’t hearts)
• So the odds of NOT getting a heart
on the first draw is 0.75.
Slide 46 of 49
Second step
(still not getting a heart on the second draw)
• Same analysis as before, except the
numbers have changed:
– The odds of NOT drawing a heart are
the number of non-hearts left in the
deck, divided by 51 (total cards
remaining)
– There are still 13 hearts, but now only
38 cards that aren’t hearts.
• So new odds of not getting a heart
are 38/51.
Slide 47 of 49
Third step
(odds of getting a heart on the third draw after
twice not getting any hearts)
• Similar, except we now are calculating
the odds of GETTING a heart.
• Still 13 hearts left, but only 50 cards in
the deck.
• So odds are 13/50.
• Total probabilities = product of all
probabilities: (3/4)(38/51)(13/50)=
(0.75)(.745)(.26)= 0.147 = 14.7% (or
about 1 out of 7 times)
Slide 48 of 49
Homework for Tuesday
1. Complete problem 17 and SHOW
WORK.
2. Problems 18-20.
3. Read Chapter 16 (it’s short, and
we’ll start it tomorrow)
4. Bring your Globe at Night data
tomorrow: we need to enter it by
Wednesday if you want extra credit.
Slide 49 of 49

similar documents