Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Honors Statistics
Friday November 21, 2014
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Daily Agenda
1. Welcome to class
2. Please find folder and take
your seat.
3. Review Homework C5#7
4. Chapter 5 Section 3 Tree diagrams
5. Homework announcement
6. Collect folders
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Chapter 5 Section 3 day 3 2014f.notebook
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OTL C5#7 homework
assignment
How do you want it - the crystal mumbo-jumbo
or statistical probability?
USE PROPER NOTATION !!!! P( ...) = page 333 ­ 334: 63 to 73 (11 problems)
Contest winners only do the ODD #'s
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#63: Get Rich page 333
Get rich A survey of 4826 randomly selected young adults (aged 19 to 25) asked, “What do you think are the chances you will have much more than a middle­class income at age 30?” The two­way table shows the responses.16 Choose a survey respondent at random.
(a) Given that the person selected is male, what’s the probability that he answered “almost certain”?
597
P(AC I M) = ­­­­­­­­­ = 0.2428 2459
(b) If the person selected said “some chance but probably not,” what’s the probability that the person is female?
426
P(F I ScPn) = ­­­­­­­­­ = 0.5983 712
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#64: A Titanic disaster page 333
319
261
627
442 765
1207
A Titanic disaster In 1912 the luxury liner Titanic, on its first voyage across the Atlantic, struck an iceberg and sank. Some passengers got off the ship in lifeboats, but many died. The two­way table gives information about adult passengers who lived and who died, by class of travel. Suppose we choose an adult passenger at random. > (a) Given that the person selected was in first class, what’s the probability that he or she survived?
197
P(Survived I FC) = ­­­­­­­­­ = 0.6176 319
> (b) If the person selected survived, what’s the probability that he or she was a third­class passenger?
151
P( third class I Survived) = ­­­­­­­­­ = 0.3416
442
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#65: Senators page 333
60
40
83
17
100
Sampling senators The two­way table describes the members of the U.S. Senate in a recent year. Suppose we select a senator at random. Consider events D: is a democrat, and F: is female.
(a) Find P(D | F). Explain what this value means.
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P(D I F) = ­­­­­­­­­ = 0.765 17
What is the probability that you are a democrat given that you are a female senator? 76.5%
(b) Find P(F | D). Explain what this value means.
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P(F I D) = ­­­­­­­­­ = 0.217 What is the probability that you are a female given that you are a democratic senator? 21.7%
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ARE THESE EVENT INDEPENDENT? ARE GENDER AND PARTY Independent events for the U.S. Senators?
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#66: Who eats breakfast ? page 333
Who eats breakfast? The following two­way table describes the 595 students who responded to a school survey about eating breakfast. Suppose we select a student at random. Consider events B: eats breakfast regularly, and M: is male.
(a) Find P(B | M). Explain what this value means.
190
P(B I M) = ­­­­­­­­­ = 0.5938 320
What is the probability that you are a male given that you eat breakfast regularly 59.38%
(b) Find P(M | B). Explain what this value means.
190
P(M I B) = ­­­­­­­­­ = 0.6333 What is the probability that you eat breakfast regularly given that you are a male? 63.33%
300
ARE THESE EVENT INDEPENDENT? ARE GENDER AND Breakfast habits Independent events?
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#67: Foreign­language study page 334
Foreign­language study Choose a student in grades 9 to 12 at random and ask if he or she is studying a language other than English. Here is the distribution of results: > (a) What’s the probability that the student is studying a language other than English?
P(other than English) = 1 = P(none) = 1 ­ 0.59 = 0.41 > (b) What is the conditional probability that a student is studying Spanish given that he or she is studying some language other than English?
0.26
P( S I other than English) = ­­­­­­­­ = 0.6341 0.41
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#68: Income tax returns page 334
Income tax returns Here is the distribution of the adjusted gross income (in thousands of dollars) reported on individual federal income tax returns in a recent year: > (a) What is the probability that a randomly chosen return shows an adjusted gross income of $50,000 or more?
P(income ≥ 50000) = 0.215 + 0.100 + 0.006 = 0.321
> (b) Given that a return shows an income of at least $50,000, what is the conditional probability that the income is at least $100,000?
0.106
P( income ≥ 100,000 I income ≥ 50,000) = ­­­­­­­­­­­ = 0.3302
0.321
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#70: Teachers and college degrees page 334
Teachers and college degrees Select an adult at random. Define events A: person has earned a college degree, and T: person’s career is teaching. Rank the following probabilities from smallest to largest. Justify your answer.
P(A) P(T) P(A | T) P(T | A)
P(T)... some of the population of adults are teachers
P(T I A) ... some people who have college degrees are teachers
P(A) ... many people have college degrees
P(A I T) ... all teachers have college degrees
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#69: Tall people and basketball players page 334
Tall people and basketball players Select an adult at random. Define events T: person is over 6 feet tall, and B: person is a professional basketball player. Rank the following probabilities from smallest to largest. Justify your answer.
P(T) P(B) P(T | B) P(B | T)
P(B) ... some people play basketball (few pro)
P(B I T) ... some tall people play basketball
P(T)... more of the population of adults are tall
P(T I B) ... almost all pro basketball players are tall
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#71: Facebook versus YouTube page 334
Facebook versus YouTube A recent survey suggests that 85% of college students have posted a profile on Facebook, 73% use YouTube regularly, and 66% do both. Suppose we select a college student at random and learn that the student has a profile on Facebook. Find the probability that the student uses YouTube regularly. Show your work. FACE N F
UTube 0.66
0.07
0.73
N U
0.19
0.08
0.27
0.85
0.15
1.00
0.66
P( UTube I Face) = ­­­­­­­­­­ = 0.7765
0.85
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#72: MAC or PC? page 334
Mac or PC? A recent census at a major university revealed that 40% of its students mainly used Macintosh computers (Macs). The rest mainly used PCs. At the time of the census, 67% of the school’s students were undergraduates. The rest were graduate students. In the census, 23% of the respondents were graduate students who said that they used PCs as their primary computers. Suppose we select a student at random from among those who were part of the census and learn that the student mainly uses a PC. Find the probability that this person is a graduate student. Show your work. MAC PC
0.30
0.37
0.67
Grad 0.10
0.23
0.33
0.40
0.60
1.00
Und
0.23
P(G I PC) = ­­­­­­­­­­ = 0.383
0.60
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#73: Free downloads? page 334
Free downloads? Illegal music downloading has become a big problem: 29% of Internet users download music files, and 67% of downloaders say they don’t care if the music is copyrighted.17 What percent of Internet users download music and don’t care if it’s copyrighted? Write the information given in terms of probabilities, and use the general multiplication rule. P( IDK and download)
P( IDK I download) = ­­­­­­­­­­­­­­­­­ P(download)
x
0.67 = ­­­­­­­­­­
0.29
x = (0.67)(0.29)
x = 0.1943
P( IDK and download) = 0.1943 or 19.4%
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Please place the following in your pocket folder:
1. your quiz and test (into folder)
2. your homework papers into binder
Please put your folders on the back shelf
at the end of the class or as directed by your teacher
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What percent of people actually have Lyme Disease,
given they have tested positive?
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Exercise C5#81: HIV testing page 335
HIV testing Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV:21 Suppose that 1% of a large population carries antibodies to HIV in their blood. We choose a person from this population at random. Given that the EIA test is positive, find the probability that the person has the antibody. Show your work.
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#81: HIV testing page 335
HIV testing Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV:21 Suppose that 1% of a large population carries antibodies to HIV in their blood. We choose a person from this population at random. Given that the EIA test is positive, find the probability that the person has the antibody. Show your work.
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
INDEPENDENCE IS NOT DISJOINT
DISJOINT SETS
could be Independent
IF two sets are independent their "picture" is intersecting.
IF a picture of two sets is intersecting the two sets may or
may not be independent.
If two sets are independent (set A and set B) then...
Knowing something about set A does nothing to help me decide if the "item" is in set B Potential independent situation ....
GIRLS BOYS
Male Staff member
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Chapter 5 Section 3 day 3 2014f.notebook
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IF two sets are independent this equation must be TRUE
P(A ∩ B) = P(A)P(B) .4 .2 .1
.3
does .2 = .6 x .3 NO
.2 ≠ .18
so sets are not independent
can you change the probabilities
and make the sets independent?
OR
Does P(A) = P(A I B) ??? 0.2
Does 0.6 = ­­­­­­­ = 0.666 NO
0.3
.5 .3 .1
.1
Let's try again ...
P(B) = P(B I A) ??
0.3
0.4 = ­­­­­­­ = 0.375 NO
0.8
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
A skips 75, 78, 91
OTL C5#8 homework
assignment
How do you want it - the crystal mumbo-jumbo
or statistical probability?
USE PROPER NOTATION !!!! P( ...) = page 334­337: 75t, 76t, 77t, 78t, 80, 89, 91, 92, 94, 96 (10 problems)
t = tree diagram DO IT!!
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#75: Box of Chocolates page 334
Box of chocolates According to Forrest Gump, “Life is like a box of chocolates. You never know what you’re gonna get.” Suppose a candy maker offers a special “Gump box” with 20 chocolate candies that look the same. In fact, 14 of the candies have soft centers and 6 have hard centers. Choose 2 of the candies from a Gump box at random.
(a) Draw a tree diagram that shows the sample space of this chance process.
13
19
soft
hard
14
20
soft
20 chocolates 14 soft 6 hard centers
6
P(soft ∩ soft)=
0.4789
P(soft ∩ hard)=
0.2211
P(hard ∩ soft)=
0.2211
P(hard ∩ hard)=
0.0789
19
14
hard
6
20
19
soft
hard
5
19
(b) Find the probability that one of the chocolates has a soft center and the other one doesn’t.
P(one soft and one hard) = 0.2211 + 0.2211 = 0.4422
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#76: Inspecting switches page 334
Inspecting switches A shipment contains 10,000 switches. Of these, 1000 are bad. An inspector draws 2 switches at random, one after the other.
(a) Draw a tree diagram that shows the sample space of this chance process.
8999
9999
9000
10000
10,000 switches
1000 bad good
P(good ∩ good)=
good
bad
P(good ∩ bad)=
1000
0.8099
0.0900
9999
9000
bad
1000
10000
9999
good
bad
999
P(bad ∩ good)=
0.0900
P(bad ∩ bad)=
0.00999
9999
(b) Find the probability that both switches are defective.
P(bad ∩ bad)=
= 0.00999
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#77: Fill'er up! page 334
Fill ‘er up! In a recent month, 88% of automobile drivers filled their vehicles with regular gasoline, 2% purchased midgrade gas, and 10% bought premium gas.18 Of those who bought regular gas, 28% paid with a credit card; of customers who bought midgrade and premium gas, 34% and 42%, respectively, paid with a credit card. Suppose we select a customer at random. (a) Draw a tree diagram to represent this situation.
.28
credit
no cre
dit
.72
20­ P(reg ∩ credit)= (0.88)(0.28) = 0.2464
P(reg ∩ no credit)=(0.88)(0.72) = 0.6336
regular .88
.34
midgrade .02
credit
no cre
dit
.66
20­ P(mid ∩ credit)= (0.02)(0.34) =0.0068
P(mid ∩ no credit)=(0.02)(0.66) = 0.0132
premium .10
.42
credit
no cre
dit
.58
20­ P(prem ∩ credit)= (0.10)(0.42) = 0.042
P(prem ∩ no credit)=(0.10)(0.58) = 0.058
(b) Find the probability that the customer paid with a credit card. Show your work.
P(credit) = 0.2464 + 0.0068 + 0.042 = 0.2952
(c) Given that the customer paid with a credit card, find the probability that she bought premium gas. Show your work.
0.042
P(premium I credit) = ­­­­­­­ = 0.142
0.2952
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#78: Urban Voters page 335
Urban voters The voters in a large city are 40% white, 40% black, and 20% Hispanic. (Hispanics may be of any race in official statistics, but here we are speaking of political blocks.) A mayoral candidate anticipates attracting 30% of the white vote, 90% of the black vote, and 50% of the Hispanic vote. Suppose we select a voter at random.
Can you tell what city elected this
candidate for Mayor?
(a) Draw a tree diagram to represent this situation.
P(White ∩ vote +) = 0.12
0.30
vote for
ite
40
0.
h
W
Black 0.40
Voters
vote again
st 0.7
Hi
sp
again
P(White ∩ vote ­) = 0.28
P(Black ∩ vote +) = 0.36
r 0.90
vote fo
vote 0
st 0.1
0
P(Black ∩ vote +) = 0.04
an
ic 0.
20
0.50
vote for
vote again
s
t 0.50
P(Hispanic ∩ vote +) = 0.10
P(Hispanic ∩ vote ­) = 0.10
(b) Find the probability that this voter votes for the mayoral candidate. Show your work.
P(votes for the mayoral candidate) = 0.12 + 0.36 + 0.10 = 0.58
(c) Given that the chosen voter plans to vote for the candidate, find the probability that the voter is black. Show your work.
0.36
P(Black voter I vote for mayoral candidate) = ­­­­­­­ = 0.62
0.58
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#80: Urban Voters page 335
Fundraising by telephone Tree diagrams can organize problems having more than two stages. The figure at top right shows probabilities for a charity calling potential donors by telephone.20 Each person called is either a recent donor, a past donor, or a new prospect. At the next stage, the person called either does or does not pledge to contribute, with conditional probabilities that depend on the donor class to which the person belongs. Finally, those who make a pledge either do or don’t actually make a contribution. Suppose we randomly select a person who is called by the charity. > a) What is the probability that the person contributed to the charity? Show your work.
> (b) Given that the person contributed, find the probability that he or she is a recent donor. Show your work.
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Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#89: Urban Voters page 335
Bright lights? A string of Christmas lights contains 20 lights. The lights are wired in series, so that if any light fails, the whole string will go dark. Each light has probability 0.02 of failing during a 3­year period. The lights fail independently of each other. Find the probability that the string of lights will remain bright for 3 years. P(lights state lit) = 1­P(one goes out) = 1­.02 = .98
but 20 lights sooooo P(lit) = .98 20 = .668
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Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#91: Urban Voters page 335
Universal blood donors People with type O­negative blood are universal donors. That is, any patient can receive a transfusion of O­negative blood. Only 7.2% of the American population have O­negative blood. If we choose 10 Americans at random who gave blood, what is the probability that at least 1 of them is a universal donor? 46
Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#92: Urban Voters page 335
Lost Internet sites Internet sites often vanish or move, so that references to them can’t be followed. In fact, 13% of Internet sites referenced in major scientific journals are lost within two years after publication.22 If we randomly select seven Internet references, from scientific journals, what is the probability that at least one of them doesn’t work two years later? 47
Chapter 5 Section 3 day 3 2014f.notebook
November 21, 2014
Exercise C5#94: Urban Voters page 335
Late flights An airline reports that 85% of its flights arrive on time. To find the probability that its next four flights into LaGuardia Airport all arrive on time, can we multiply (0.85)(0.85)(0.85)(0.85)? Why or why not? 48
Chapter 5 Section 3 day 3 2014f.notebook
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Exercise C5#96: Urban Voters page 335
The probability of a flush A poker player holds a flush when all 5 cards in the hand belong to the same suit. We will find the probability of a flush when 5 cards are dealt. Remember that a deck contains 52 cards, 13 of each suit, and that when the deck is well shuffled, each card dealt is equally likely to be any of those that remain in the deck.
(a) We will concentrate on spades. What is the probability that the first card dealt is a spade? What is the conditional probability that the second card is a spade given that the first is a spade?
P(spade) = 13/52 = .25
P(spade l spade) = (12/51) = .23529
(b) Continue to count the remaining cards to find the conditional probabilities of a spade on the third, the fourth, and the fifth card given in each case that all previous cards are spades.
P(spade l spade l spade) = (11/50) = .22
P(spade l spade l spade l spade) = (10/49) =.2041
P(spade l spade l spade l spade l spade) = (9/48) = .1875
(c) The probability of being dealt 5 spades is the product of the five probabilities you have found. Why? What is this probability?
using the extended multiplication rule gives the probability of a flush
P(spade flush) = (.25)(.2353)(.22)(.2041)(.1875) = .0004952
(d) The probability of being dealt 5 hearts or 5 diamonds or 5 clubs is the same as the probability of being dealt 5 spades. What is the probability of being dealt a flush?
multiply answer c by 4 (4)(.0004952) = .001981
about 2 out of 1000 hands (in the looooong run)
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