Chapter 3 Extra Practice

3.1 Introduction and Visualizing Bivariate Categorical Data

1. The following figure shows a random sample of 100 hikers and the areas of hiking they prefer.

Figure 3.19: Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___

a. Complete the table.

a.

Figure 3.20: Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 11 45
Male 16 25 14 55
Total 34 41 25 100

b. Are the events “being female” and “preferring the coastline” independent events?

Let F = being female and let C = preferring the coastline.

  1. Find P(F AND C).
  2. Find P(F)P(C)

Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.

b.

  1. P(F AND C) = \frac{18}{100} = 0.18
  2. P(F)P(C) = \left(\frac{45}{100}\right)\left(\frac{34}{100}\right) = (0.45)(0.34) = 0.153

P(F AND C) ≠ P(F)P(C), so the events F and C are not independent.

 

c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.

  1. What word tells you this is a conditional?
  2. Fill in the blanks and calculate the probability: P(___|___) = ___.
  3. Is the sample space for this problem all 100 hikers? If not, what is it?

c.The word ‘given’ tells you that this is a conditional.

  1. P(M|L) = \frac{25}{41}
  2. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.

  1. Find P(F).
  2. Find P(P).
  3. Find P(F AND P).
  4. Find P(F OR P).

d.

  1. P(F) = \frac{45}{100}
  2. P(P) = \frac{25}{100}
  3. P(F AND P) = \frac{11}{100}
  4. P(F OR P) = \frac{45}{100} + \frac{25}{100} – \frac{11}{100} = \frac{59}{100}

2. The figure below shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.

Figure 3.21
Gender Lake Path Hilly Path Wooded Path Total
Female 45 38 27 110
Male 26 52 12 90
Total 71 90 39 200
  1. Out of the males, what is the probability that the cyclist prefers a hilly path?
  2. Are the events “being male” and “preferring the hilly path” independent events?

3. Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is \frac{1}{5}\text{} and the probability he is not caught is \frac{4}{5}\text{}. If he goes out the second door, the probability he gets caught by Alissa is \frac{1}{4} and the probability he is not caught is \frac{3}{4}. The probability that Alissa catches Muddy coming out of the third door is \frac{1}{2} and the probability she does not catch Muddy is \frac{1}{2}. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is \frac{1}{3}.

Figure 3.22: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} ____
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} ____
Total ____ ____ ____ 1
  • The first entry \frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Caught)
  • The entry \frac{4}{15} =\left(\frac{4}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Not Caught)

Verify the remaining entries.

a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

a.

Figure 3.23: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} \frac{19}{60}
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} \frac{41}{60}
Total \frac{5}{15} \frac{4}{12} \frac{2}{6} 1

b. What is the probability that Alissa does not catch Muddy?

b. \frac{41}{60}

c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

c. \frac{9}{19}

4. The figure below relates the weights and heights of a group of individuals participating in an observational study.

Figure 3.24
Weight/Height Tall Medium Short Totals
Obese 18 28 14
Normal 20 51 28
Underweight 12 25 9
Totals
  1. Find the total for each row and column
  2. Find the probability that a randomly chosen individual from this group is Tall.
  3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
  4. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Obese.
  5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
  6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
  7. Are the events Obese and Tall independent?

5. There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilities that have multiple dependent variables.

Use the following information to answer the next four exercises. The figure below shows a random sample of musicians and how they learned to play their instruments.

Figure 3.25
Gender Self-taught Studied in School Private Instruction Total
Female 12 38 22 72
Male 19 24 15 58
Total 31 62 37 130

a. Find P(musician is a female).

b. Find P(musician is a male AND had private instruction).

  • P(musician is a male AND had private instruction) = \frac{15}{130} = \frac{3}{26} = 0.12

c. Find P(musician is a female OR is self taught).

d. Are the events “being a female musician” and “learning music in school” mutually exclusive events?

  • P(being a female musician AND learning music in school) = \frac{38}{130} = \frac{19}{65}= 0.29
  • No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school).

6. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites.[1]

Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

Figure 3.26: Smoking Levels by Ethnicity
Smoking Level African American Native Hawaiian Latino Japanese Americans White TOTALS
1–10
11–20
21–30
31+
TOTALS

a. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

  • \frac{35,065}{100,450}

b. Find the probability that the person was Latino.

c. In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.

  • To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is \frac{4,715}{100,450}.

d. In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.

e. In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.

  • To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is \frac{4715}{15,273}.

f. Prove that smoking level/day and ethnicity are dependent events.

7. The figure below contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S. [2]

Figure 3.27: United States Crime Index Rates Per 100,000 Inhabitants 2008–2011
Year Robbery Burglary Rape Vehicle Total
2008 145.7 732.1 29.7 314.7
2009 133.1 717.7 29.1 259.2
2010 119.3 701 27.7 239.1
2011 113.7 702.2 26.8 229.6
Total

TOTAL each column and each row. Total data = 4,520.7

  1. Find P (2009 AND Robbery).
  2. Find P (2010 AND Burglary).
  3. Find P (2010 OR Burglary).
  4. Find P (2011|Rape).
  5. Find P (Vehicle|2008).

8. The following figure shows a random sample of 100 hikers and the areas of hiking they prefer.

Figure 3.28: Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___

a. Complete the table.

b. Are the events “being female” and “preferring the coastline” independent events? Let F = being female and let C = preferring the coastline.

  • Find P(F AND C).
  • Find P(F)P(C)
  • Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.

c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.

  1. What word tells you this is a conditional?
  2. Fill in the blanks and calculate the probability: P(___|___) = ___.
  3. Is the sample space for this problem all 100 hikers? If not, what is it?

d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.

  1. Find P(F).
  2. Find P(P).
  3. Find P(F AND P).
  4. Find P(F OR P).

Solutions:

a.

Figure 3.29: Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 11 45
Male 16 25 14 55
Total 34 41 25 100

b.

  • P(F AND C) = \frac{18}{100} = 0.18
  • P(F)P(C) = \left(\frac{45}{100}\right)\left(\frac{34}{100}\right) = (0.45)(0.34) = 0.153
  • P(F AND C) ≠ P(F)P(C), so the events F and C are not independent.

c.

  • The word ‘given’ tells you that this is a conditional.
  • P(M|L) = \frac{25}{41}
  • No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

d.

  • P(F) = \frac{45}{100}
  • P(P) = \frac{25}{100}
  • P(F AND P) = \frac{11}{100}
  • P(F OR P) = \frac{45}{100} + \frac{25}{100} – \frac{11}{100} = \frac{59}{100}

9. The figure below shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.

Figure 3.30
Gender Lake Path Hilly Path Wooded Path Total
Female 45 38 27 110
Male 26 52 12 90
Total 71 90 39 200
  1. Out of the males, what is the probability that the cyclist prefers a hilly path?
  2. Are the events “being male” and “preferring the hilly path” independent events?

10. Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is \frac{1}{5}\text{} and the probability he is not caught is \frac{4}{5}\text{}. If he goes out the second door, the probability he gets caught by Alissa is \frac{1}{4} and the probability he is not caught is \frac{3}{4}. The probability that Alissa catches Muddy coming out of the third door is \frac{1}{2} and the probability she does not catch Muddy is \frac{1}{2}. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is \frac{1}{3}.

Figure 3.31: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} ____
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} ____
Total ____ ____ ____ 1
  • The first entry \frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Caught)
  • The entry \frac{4}{15} =\left(\frac{4}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Not Caught)

Verify the remaining entries.

 

a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

a.

Figure 3.32: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} \frac{19}{60}
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} \frac{41}{60}
Total \frac{5}{15} \frac{4}{12} \frac{2}{6} 1

b. What is the probability that Alissa does not catch Muddy?

b. \frac{41}{60}

c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

c. \frac{9}{19}

11. The figure below relates the weights and heights of a group of individuals participating in an observational study.

Figure 3.33
Weight/Height Tall Medium Short Totals
Obese 18 28 14
Normal 20 51 28
Underweight 12 25 9
Totals
  1. Find the total for each row and column
  2. Find the probability that a randomly chosen individual from this group is Tall.
  3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
  4. Find the probability that a randomly chosen individual from this group is Tall given that the individual is Obese.
  5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
  6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
  7. Are the events Obese and Tall independent?

12. Use the following information to answer the next four exercises. The figure below shows a random sample of musicians and how they learned to play their instruments.

Figure 3.34
Gender Self-taught Studied in School Private Instruction Total
Female 12 38 22 72
Male 19 24 15 58
Total 31 62 37 130

Find P(musician is a female).

Find P(musician is a male AND had private instruction).

P(musician is a male AND had private instruction) = \frac{15}{130} = \frac{3}{26} = 0.12

Find P(musician is a female OR is self taught).

Are the events “being a female musician” and “learning music in school” mutually exclusive events?

P(being a female musician AND learning music in school) = \frac{38}{130} = \frac{19}{65}= 0.29

No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school).

13. Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites. [3]

Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

Figure 3.35: Smoking Levels by Ethnicity
Smoking Level African American Native Hawaiian Latino Japanese Americans White TOTALS
1–10
11–20
21–30
31+
TOTALS

Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

\frac{35,065}{100,450}

Find the probability that the person was Latino.

In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.

To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is \frac{4,715}{100,450}.

In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.

In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.

To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is \frac{4715}{15,273}.

Prove that smoking level/day and ethnicity are dependent events.

14. In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.
Total = 64 + 24 + 24 + 9 = 121

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.
Figure 3.36

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:

R1R1R1R2R1R3R2R1R2R2R2R3R3R1R3R2R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

a. List the 24 BR outcomes: B1R1, B1R2, B1R3, …

  • B1R1B1R2B1R3B2R1B2R2B2R3B3R1B3R2B3R3B4R1B4R2B4R3B5R1B5R2B5R3B6R1B6R2B6R3B7R1B7R2B7R3B8R1B8R2B8R3

b. Using the tree diagram, calculate P(RR).

  • P(RR) =\left(\frac{3}{11}\right)\left(\frac{3}{11}\right) =\frac{9}{121}

c. Using the tree diagram, calculate P(RB OR BR).

  • P(RB OR BR) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right)\) + \(\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)=\frac{48}{121}

d. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).

  • P(R on 1st draw AND B on 2nd draw) = P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) = \frac{24}{121}

e. Using the tree diagram, calculate P(R on 2nd draw GIVEN B on 1st draw).

  • P(R on 2nd draw GIVEN B on 1st draw) = P(R on 2nd|B on 1st) = \frac{24}{88} = \frac{3}{11}

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. \frac{24}{88} = \frac{3}{11}.

f. Using the tree diagram, calculate P(BB).

  • P(BB) = \frac{64}{121}

g. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).

  • P(B on 2nd draw|R on 1st draw) = \frac{8}{11}

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then \frac{24}{33}.

15. An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. “Without replacement” means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example:

\left(\frac{3}{11}\right)\left(\frac{2}{10}\right) = \frac{6}{110}.
Total = \frac{56+24+24+6}{110} = \frac{110}{110}=1

This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.
Figure 3.37

NOTE: If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.

Calculate the following probabilities using the tree diagram.

 

a. P(RR) = ________

  • P(RR) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}

b. Fill in the blanks: P(RB OR BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+ (___)(___)= \frac{48}{110}

  • P(RB OR BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{48}{110}

c. P(R on 2nd|B on 1st) =

  • \frac{3}{10}

d. Fill in the blanks. P(R on 1st AND B on 2nd) = P(RB) = (___)(___) = \frac{24}{100}

  • P(R on 1st AND B on 2nd) = P(RB) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{100}

e. Find P(BB).

  • P(BB) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right)

f. Find P(B on 2nd|R on 1st).

  • Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = \frac{8}{10}

If we are using probabilities, we can label the tree in the following general way.

This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).
Figure 3.38
  • P(R|R) here means P(R on 2nd|R on 1st)
  • P(B|R) here means P(B on 2nd|R on 1st)
  • P(R|B) here means P(R on 2nd|B on 1st)
  • P(B|B) here means P(B on 2nd|B on 1st)

16. In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 12F and 40N. The second branch has a set of two lines (12F and 40N) for each line of the first branch. Multiply along each line to find 144FF, 480FN, 480NF, and 1,600NN.
Figure 3.39

17. In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows 2 lines: F 12/52 and N 40/52. The second branch has a set of 2 lines (F 11/52 and N 40/51) for each line of the first branch. Multiply along each line to find FF 121/2652, FN 480/2652, NF 480/2652, and NN 1560/2652.
Figure 3.40
  1. Find P(FN OR NF).
  2. Find P(N|F).
  3. Find P(at most one face card).
    Hint: “At most one face card” means zero or one face card.
  4. Find P(at least on face card).
    Hint: “At least one face card” means one or two face cards.

18. A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

This is a tree diagram with branches showing probabilities of kitten choices. The first branch shows two lines: T 4/9 and B 5/9. The second branch has a set of 2 lines for each first branch line. Below T 4/9 are T 3/8 and B 5/8. Below B 5/9 are T 4/8 and B 4/8. Multiply along each line to find probabilities of possible combinations.
Figure 3.41
  1. What is the probability that both kittens are tabby?

a.\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)

b.\left(\frac{4}{9}\right)\left(\frac{4}{9}\right)

c.\left(\frac{4}{9}\right)\left(\frac{3}{8}\right)

d.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)

2. What is the probability that one kitten of each coloring is selected?

a.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)

b.\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)

c.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)

d.\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)

3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?

4. What is the probability of choosing two kittens of the same color?

Solutions: 1. c, 2. d, 3. \frac{4}{8}, 4. \frac{32}{72}

19. Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

20. Flip two fair coins. Let A = tails on the first coin. Let B = tails on the second coin. Then A = {TT, TH} and B = {TT, HT}. Therefore, A AND B = {TT}. A OR B = {TH, TT, HT}.

The sample space when you flip two fair coins is X = {HH, HT, TH, TT}. The outcome HH is in NEITHER A NOR B. Draw a Venn Diagram.

Solution:

This is a venn diagram. An oval representing set A contains Tails + Heads and Tails + Tails. An oval representing set B also contains Tails + Tails, along with Heads + Tails. The universe S contains Heads + Heads, but this value is not contained in either set A or B.
Figure 3.42

21. Roll a fair, six-sided die. Let A = a prime number of dots is rolled. Let B = an odd number of dots is rolled. Then A = {2, 3, 5} and B = {1, 3, 5}. Therefore, A AND B = {3, 5}. A OR B = {1, 2, 3, 5}. The sample space for rolling a fair die is S = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

22. Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event C = {green, blue, purple} and event P = {red, yellow, blue}. Then C AND P = {blue} and C OR P = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.

23. Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let W = works a second job and S = spouse also works.

24. A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood. [4] [5]

This is an empty Venn diagram showing two overlapping circles. The left circle is labeled O and the right circle is labeled RH-.
Figure 3.43

The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.

We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let O = African American with Type O blood and R = African American with Rh- factor.

  1. P(O) = ___________
  2. P(R) = ___________
  3. P(O AND R) = ___________
  4. P(O OR R) = ____________
  5. In the Venn Diagram, describe the overlapping area using a complete sentence.
  6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

a. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.

25. In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.

  1. Draw a Venn diagram representing the situation.
  2. Find the probability that the customer buys either a novel or anon-fiction book.
  3. In the Venn diagram, describe the overlapping area using a complete sentence.
  4. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.

26. The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: C = a man develops cancer in his lifetime; P = man has at least one false positive. Construct a tree diagram of the situation.

[6]

This is a tree diagram with two branches. The first branch, labeled Cancer, shows two lines: 0.4567 C and 0.5433 C'. The second branch is labeled False Positive. From C, there are two lines: 0 P and 1 P'. From C', there are two lines: 0.51 P and 0.49 P'.
Figure 3.44

27. This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (R), four yellow (Y) and five blue (B) beads. For the coin, P(H) = \frac{2}{3} and P(T) = \frac{1}{3} where H is heads and T is tails.

Tree diagram with 2 branches. The first branch consists of 2 lines of H=2/3 and T=1/3. The second branch consists of 2 sets of 3 lines each with the both sets containing R=3/12, Y=4/12, and B=5/12.
Figure 3.45

Find P(tossing a Head on the coin AND a Red bead)

  1. \frac{2}{3}
  2. \frac{5}{15}
  3. \frac{6}{36}
  4. \frac{5}{36}

Find P(Blue bead).

  1. \frac{15}{36}
  2. \frac{10}{36}
  3. \frac{10}{12}
  4. \frac{6}{36}

28. A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?)

  1. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.
  2. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain.
  3. For each complete path through the tree, write the event it represents and find the probabilities.
  4. Let S be the event that both cookies selected were the same flavor. Find P(S).
  5. Let T be the event that the cookies selected were different flavors. Find P(T) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.
  6. Let U be the event that the second cookie selected is a butter cookie. Find P(U).

29. Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.

Suppose that you randomly draw two cards, one at a time, with replacement.
Let G1 = first card is green
Let G2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P(G1 AND G2).
  3. Find P(at least one green).
  4. Find P(G2|G1).
  5. Are G2 and G1 independent events? Explain why or why not.
  1. This is a tree diagram with branches showing probabilities of each draw. The first branch shows two lines: 5/8 Green and 3/8 Yellow. The second branch has a set of two lines (5/8 Green and 3/8 Yellow) for each line of the first branch.
    Figure 3.46
  2. P(GG) = \left(\frac{5}{8}\right)\left(\frac{5}{8}\right) = \frac{25}{64}
  3. P(at least one green) = P(GG) + P(GY) + P(YG) = \frac{25}{64} + \frac{15}{64} + \frac{15}{64} = \frac{55}{64}
  4. P(G|G) = \frac{5}{8}
  5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn; the composition of cards in the bag remains the same from draw one to draw two.

Suppose that you randomly draw two cards, one at a time, without replacement.
G1 = first card is green
G2 = second card is green

  1. Draw a tree diagram of the situation.
  2. Find P(G1 AND G2).
  3. Find P(at least one green).
  4. Find P(G2|G1).
  5. Are G2 and G1 independent events? Explain why or why not.

30. The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over. [7]

Complete the following.

  1. Construct a table or a tree diagram of the situation.
  2. Find P(driver is female).
  3. Find P(driver is age 65 or over|driver is female).
  4. Find P(driver is age 65 or over AND female).
  5. In words, explain the difference between the probabilities in part c and part d.
  6. Find P(driver is age 65 or over).
  7. Are being age 65 or over and being female mutually exclusive events? How do you know?
  1. Figure 3.47
    <20 20–64 >64 Totals
    Female 0.0244 0.3954 0.0661 0.486
    Male 0.0259 0.4186 0.0695 0.514
    Totals 0.0503 0.8140 0.1356 1
  2. P(F) = 0.486
  3. P(>64|F) = 0.1361
  4. P(>64 and F) = P(F) P(>64|F) = (0.486)(0.1361) = 0.0661
  5. P(>64|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older.
  6. P(>64) = P(>64 and F) + P(>64 and M) = 0.1356
  7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) = 0.0661.

Suppose that 10,000 U.S. licensed drivers are randomly selected.

  1. How many would you expect to be male?
  2. Using the table or tree diagram, construct a contingency table of gender versus age group.
  3. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.

31. Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation. [8]

  1. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.
  2. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?
  3. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work?
  4. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool?
  1. Figure 3.48
    Car, Truck or Van Walk Public Transportation Other Totals
    Alone 0.7318
    Not Alone 0.1332
    Totals 0.8650 0.0390 0.0530 0.0430 1
  2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = 0.7318 + 0.0390 = 0.7708.
  3. Make the same assumptions as in (b) we have: (0.7708)(1,000) = 771
  4. (0.1332)(1,000) = 133

32. When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event H) while 110 showed a tail (event T). On that basis, they claimed that it is not a fair coin.

  1. Based on the given data, find P(H) and P(T).
  2. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.
  3. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.
  4. Use the tree to find the probability of obtaining at least one head.

33. The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories: [9]

Figure 3.49: AIDS statistics * includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 ____
Male 2,146 463 60 135 ____
Totals ____ ____ ____ ____ ____

Suppose a person with AIDS in Santa Clara County is randomly selected.

  1. Find P(Person is female).
  2. Find P(Person has a risk factor heterosexual contact).
  3. Find P(Person is female OR has a risk factor of IV drug user).
  4. Find P(Person is female AND has a risk factor of homosexual/bisexual).
  5. Find P(Person is male AND has a risk factor of IV drug user).
  6. Find P(Person is female GIVEN person got the disease from heterosexual contact).
  7. Construct a Venn diagram. Make one group females and the other group heterosexual contact.

The completed contingency table is as follows:

Figure 3.50: AIDS statistics solution * includes homosexual/bisexual IV drug users
Homosexual/Bisexual IV Drug User* Heterosexual Contact Other Totals
Female 0 70 136 49 255
Male 2,146 463 60 135 2,804
Totals 2,146 533 196 184 3,059
  1. \frac{255}{3059}
  2. \frac{196}{3059}
  3. \frac{718}{3059}
  4. 0
  5. \frac{463}{3059}
  6. \frac{136}{196}
  7. Figure 3.51

Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact.

  1. Find P(Person is female).
  2. Find P(Person obtained the disease through heterosexual contact).
  3. Find P(Person is female GIVEN person got the disease from heterosexual contact)
  4. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.

34. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection. [10]

Figure 3.52
Up for reelection: Democratic Party Republican Party Other Total
November 2014 20 13 0
November 2016 10 24 0
Total

a. What is the probability that a randomly selected senator has an “Other” affiliation?

  • 0

b. What is the probability that a randomly selected senator is up for reelection in November 2016?

c. What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?

  • \frac{10}{67}

d. What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?

e. Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?

  • \frac{10}{34}

f. Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?

g. The events “Republican” and “Up for reelection in 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

h. The events “Other” and “Up for reelection in November 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

35. The figure below gives the number of suicides estimated in the U.S. for a recent year by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population.

Figure 3.53
Race and Sex 1–14 15–24 25–64 over 64 TOTALS
white, male 210 3,360 13,610 22,050
white, female 80 580 3,380 4,930
black, male 10 460 1,060 1,670
black, female 0 40 270 330
all others
TOTALS 310 4,650 18,780 29,760

Do not include “all others” for parts f and g.

  1. Fill in the column for the suicides for individuals over age 64.
  2. Fill in the row for all other races.
  3. Find the probability that a randomly selected individual was a white male.
  4. Find the probability that a randomly selected individual was a black female.
  5. Find the probability that a randomly selected individual was black
  6. Find the probability that a randomly selected individual was a black or white male.
  7. Out of the individuals over age 64, find the probability that a randomly selected individual was a black or white male.
  1. Figure 3.54
    Race and Sex 1–14 15–24 25–64 over 64 TOTALS
    white, male 210 3,360 13,610 4,870 22,050
    white, female 80 580 3,380 890 4,930
    black, male 10 460 1,060 140 1,670
    black, female 0 40 270 20 330
    all others 100
    TOTALS 310 4,650 18,780 6,020 29,760
  2. Figure 3.55
    Race and Sex 1–14 15–24 25–64 over 64 TOTALS
    white, male 210 3,360 13,610 4,870 22,050
    white, female 80 580 3,380 890 4,930
    black, male 10 460 1,060 140 1,670
    black, female 0 40 270 20 330
    all others 10 210 460 100 780
    TOTALS 310 4,650 18,780 6,020 29,760
  3. \frac{\text{22,050}}{\text{29,760}}
  4. \frac{\text{330}}{\text{29,760}}
  5. \frac{\text{2,000}}{\text{29,760}}
  6. \frac{23720}{\left(29760-780\right)} = \frac{23720}{28980}
  7. \frac{5010}{\left(6020-100\right)} = \frac{5010}{5920}

36. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected. [11]

Figure 3.56
NAME Single Double Triple Home Run TOTAL HITS
Babe Ruth 1,517 506 136 714 2,873
Jackie Robinson 1,054 273 54 137 1,518
Ty Cobb 3,603 174 295 114 4,189
Hank Aaron 2,294 624 98 755 3,771
TOTAL 8,471 1,577 583 1,720 12,351

Find P(hit was made by Babe Ruth).

  1. \frac{1518}{2873}
  2. \frac{2873}{12351}
  3. \frac{583}{12351}
  4. \frac{4189}{12351}

Find P(hit was made by Ty Cobb|The hit was a Home Run).

  1. \frac{4189}{12351}
  2. \frac{114}{1720}
  3. \frac{1720}{4189}
  4. \frac{114}{12351}

37. The figure below identifies a group of children by one of four hair colors, and by type of hair.

Figure 3.57
Hair Type Brown Blond Black Red Totals
Wavy 20 15 3 43
Straight 80 15 12
Totals 20 215
  1. Complete the table.
  2. What is the probability that a randomly selected child will have wavy hair?
  3. What is the probability that a randomly selected child will have either brown or blond hair?
  4. What is the probability that a randomly selected child will have wavy brown hair?
  5. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
  6. If B is the event of a child having brown hair, find the probability of the complement of B.
  7. In words, what does the complement of B represent?

38. In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.[12]

Figure 3.58
Shirt# ≤ 210 211–250 251–290 > 290
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

  1. Find the probability that his shirt number is from 1 to 33.
  2. Find the probability that he weighs at most 210 pounds.
  3. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
  4. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
  5. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
  1. \frac{26}{106}
  2. \frac{33}{106}
  3. \frac{21}{106}
  4. \left(\frac{26}{106}\right)\) + \(\left(\frac{33}{106}\right)\) - \(\left(\frac{21}{106}\right)\) = \(\left(\frac{38}{106}\right}
  5. \frac{21}{33}

3.2 Introduction to Bi-variate Data and Scatterplots

1. The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. The figure below shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data.

Figure 9.19
Year Cuba’s PPP Year Cuba’s PPP
1999 1,700 2006 4,000
2000 1,700 2007 11,000
2002 2,300 2008 9,500
2003 2,900 2009 9,700
2004 3,000 2010 9,900
2005 3,500

2. The following table shows the poverty rates and cell phone usage in the United States. Construct a scatter plot of the data.

Figure 9.20
Year Poverty Rate Cellular Usage per Capita
2003 12.7 54.67
2005 12.6 74.19
2007 12 84.86
2009 12 90.82

3. Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data. Note that tuition is the independent variable and salary is the dependent variable.

Figure 9.21
School Mid-Career Salary (in thousands) Yearly Tuition
Princeton 137 28,540
Harvey Mudd 135 40,133
CalTech 127 39,900
US Naval Academy 122 0
West Point 120 0
MIT 118 42,050
Lehigh University 118 43,220
NYU-Poly 117 39,565
Babson College 117 40,400
Stanford 114 54,506

 

3.2 Measures of Association

1. Can a coefficient of determination be negative? Why or why not?

2. The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. The figure below shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data.

Figure 9.19
Year Cuba’s PPP Year Cuba’s PPP
1999 1,700 2006 4,000
2000 1,700 2007 11,000
2002 2,300 2008 9,500
2003 2,900 2009 9,700
2004 3,000 2010 9,900
2005 3,500

Find:

  • r
  • r2x

3. The following table shows the poverty rates and cell phone usage in the United States. Construct a scatter plot of the data.

Figure 9.20
Year Poverty Rate Cellular Usage per Capita
2003 12.7 54.67
2005 12.6 74.19
2007 12 84.86
2009 12 90.82

Find:

  • r
  • r2

4. Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data. Note that tuition is the independent variable and salary is the dependent variable.

Figure 9.21
School Mid-Career Salary (in thousands) Yearly Tuition
Princeton 137 28,540
Harvey Mudd 135 40,133
CalTech 127 39,900
US Naval Academy 122 0
West Point 120 0
MIT 118 42,050
Lehigh University 118 43,220
NYU-Poly 117 39,565
Babson College 117 40,400
Stanford 114 54,506

Find:

  • r
  • r2

3.3 Modeling Linear Relationships

1.  A random sample of ten professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars).

Figure 9.22
x y x y
0 2 5 12
3 8 4 9
2 7 3 9
1 3 0 3
5 13 4 10

a. Draw a scatter plot of the data.

b. Use regression to find the equation for the line of best fit.

  • ŷ = 2.23 + 1.99x

c. Draw the line of best fit on the scatter plot.

d. What is the slope of the line of best fit? What does it represent?

  • The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another \$1.99 million in pay each year.

e. What is the y-intercept of the line of best fit? What does it represent?

f. What does an r value of zero mean?

  • It means that there is no correlation between the data sets.

g. When n = 2 and r = 1, are the data significant? Explain.

h. When n = 100 and r = -0.89, is there a significant correlation? Explain.

  • Yes, there are enough data points and the value of r is strong enough to show that there is a strong negative correlation between the data sets.

2. What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern?

3.4 Cautions about Regression

1. The following table shows economic development measured in per capita income PCINC.

Figure 9.23
Year PCINC Year PCINC
1870 340 1920 1050
1880 499 1930 1170
1890 592 1940 1364
1900 757 1950 1836
1910 927 1960 2132
  1. What are the independent and dependent variables?
  2. Draw a scatter plot.
  3. Use regression to find the line of best fit and the correlation coefficient.
  4. Interpret the significance of the correlation coefficient.
  5. Is there a linear relationship between the variables?
  6. Find the coefficient of determination and interpret it.
  7. What is the slope of the regression equation? What does it mean?
  8. Use the line of best fit to estimate PCINC for 1900, for 2000.
  9. Determine if there are any outliers.

2. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69.

Scatterplot that shows a positive, very linear pattern of dots except for one (6, 60). A line of best fit is overlayed on the graph that closely follows the linear dots.
Figure 9.24

a. Do there appear to be any outliers?

  • Yes, there appears to be an outlier at (6, 58).

b. A point is removed, and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why?

c. What effect did the potential outlier have on the line of best fit?

  • The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate is a predictor for the data.

d. Are you more or less confident in the predictive ability of the new line of best fit?

e. The Sum of Squared Errors for a data set of 18 numbers is 49. What is the standard deviation?

  • s = 1.75

f. The Standard Deviation for the Sum of Squared Errors for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier?

3. The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level).

Figure 9.25
Height (in feet) Stories
1,050 57
428 28
362 26
529 40
790 60
401 22
380 38
1,454 110
1,127 100
700 46
  1. Using “stories” as the independent variable and “height” as the dependent variable, make a scatter plot of the data.
  2. Does it appear from inspection that there is a relationship between the variables?
  3. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx
  4. Find the correlation coefficient. Is it significant?
  5. Find the estimated heights for 32 stories and for 94 stories.
  6. Based on the data, is there a linear relationship between the number of stories in tall buildings and the height of the buildings?
  7. Are there any outliers in the data? If so, which point(s)?
  8. What is the estimated height of a building with six stories? Does the least squares line give an accurate estimate of height? Explain why or why not.
  9. Based on the least squares line, adding an extra story is predicted to add about how many feet to a building?
  10. What is the slope of the least squares (best-fit) line? Interpret the slope.

4. Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration.

Percent return: 74, 66, 81, 52, 73, 62, 52, 45, 62, 46, 60, 46, 38
Percent new: 5, 6, 8, 11, 12, 15, 16, 17, 18, 18, 19, 20, 20

  1. Enter the data into your calculator and make a scatter plot.
  2. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
  3. Explain in words what the slope and y-intercept of the regression line tell us.
  4. How well does the regression line fit the data? Explain your response.
  5. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
  6. An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70% of the adults from the previous year have returned. What is the prediction?
  • Solution to a and b: Check student’s solution.
  • Solution to c: The slope of the regression line is -0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32% new sparrow hawks, which doesn’t make sense since if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3%.
  • Solution to d: If we examine r2, we see that only 57.52% of the variation in the percent of new birds is explained by the model and the correlation coefficient, r = –.7584 only indicates a somewhat strong correlation between returning and new percentages.
  • Solution to e: The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66%, our observed new percentage, 6%, is almost 6% less than the predicted new value of 11.98%. If we remove this data pair, we see only an adjusted slope of -.2789 and an adjusted intercept of 30.9816. In other words, even though this data generates the largest residual, it is not an outlier, nor is the data pair an influential point.
  • Solution to f: If there are 70% returning birds, we would expect to see y = –.2789(70) + 30.9816 = 0.114 or 11.4% new birds in the colony.

5. The following table shows data on average per capita coffee consumption and heart disease rate in a random sample of 10 countries.

Figure 9.26
Yearly coffee consumption in liters 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7
Death from heart diseases 221 167 131 191 220 297 71 172 211 300
  1. Enter the data into your calculator and make a scatter plot.
  2. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
  3. Explain in words what the slope and y-intercept of the regression line tell us.
  4. How well does the regression line fit the data? Explain your response.
  5. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
  6. Do the data provide convincing evidence that there is a linear relationship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.

6. The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days:

Figure 9.27
Swim Time Heart Rate
34.12 144
35.72 152
34.72 124
34.05 140
34.13 152
35.73 146
36.17 128
35.57 136
35.37 144
35.57 148
  1. Enter the data into your calculator and make a scatter plot.
  2. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a.
  3. Explain in words what the slope and y-intercept of the regression line tell us.
  4. How well does the regression line fit the data? Explain your response.
  5. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
  • Check student’s solution.
  • Check student’s solution.
  • We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate will decrease by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. While the slope has meaning (the longer it takes to swim 2,000 meters, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low.
  • Since only 1.5% of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship.
  • The point (34.72, 124) generates the largest residual of –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes –2.953 with the y-intercept changing to 247.1616. While the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y-intercept becomes more meaningful.

7. A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are white males.

Figure 9.28
Population Size Homicide rate per 100,000 people
558,724 8.6
538,584 8.9
519,171 8.52
500,457 8.89
482,418 13.07
465,029 14.57
448,267 21.36
432,109 28.03
416,533 31.49
401,518 37.39
387,046 46.26
373,095 47.24
359,647 52.33
  1. Use your calculator to construct a scatter plot of the data. What should the independent variable be? Why?
  2. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot.
  3. Discuss what the following mean in context.
    1. The slope of the regression equation
    2. The y-intercept of the regression equation
    3. The correlation r
    4. The coefficient of determination r2.
  4. Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.

8. Use the table below to answer (a) and (b).

Figure 9.29
School Mid-Career Salary (in thousands) Yearly Tuition
Princeton 137 28,540
Harvey Mudd 135 40,133
CalTech 127 39,900
US Naval Academy 122 0
West Point 120 0
MIT 118 42,050
Lehigh University 118 43,220
NYU-Poly 117 39,565
Babson College 117 40,400
Stanford 114 54,506

a. Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer.

b. If we remove the two service academies (the tuition is \$0.00), we construct a new regression equation of y = –0.0009x + 160 with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set.

9. The average number of people in a family that attended college for various years is given below.

Figure 9.30
Year Number of Family Members Attending College
1969 4.0
1973 3.6
1975 3.2
1979 3.0
1983 3.0
1988 3.0
1991 2.9
  1. Using “year” as the independent variable and “Number of Family Members Attending College” as the dependent variable, draw a scatter plot of the data.
  2. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  3. Find the correlation coefficient. Is it significant?
  4. Pick two years between 1969 and 1991 and find the estimated number of family members attending college.
  5. Based on the data, is there a linear relationship between the year and the average number of family members attending college?
  6. Using the least-squares line, estimate the number of family members attending college for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not.
  7. Are there any outliers in the data?
  8. What is the estimated average number of family members attending college for 1986? Does the least squares line give an accurate estimate for that year? Explain why or why not.
  9. What is the slope of the least squares (best-fit) line? Interpret the slope.

10. The percent of female wage and salary workers who are paid hourly rates is given in below for the years 1979 to 1992.

Figure 9.31
Year Percent of workers paid hourly rates
1979 61.2
1980 60.7
1981 61.3
1982 61.3
1983 61.8
1984 61.7
1985 61.8
1986 62.0
1987 62.7
1990 62.8
1992 62.9
  1. Using “year” as the independent variable and “percent” as the dependent variable, draw a scatter plot of the data.
  2. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  3. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  4. Find the correlation coefficient. Is it significant?
  5. Find the estimated percents for 1991 and 1988.
  6. Based on the data, is there a linear relationship between the year and the percent of female wage and salary earners who are paid hourly rates?
  7. Are there any outliers in the data?
  8. What is the estimated percent for the year 2050? Does the least-squares line give an accurate estimate for that year? Explain why or why not.
  9. What is the slope of the least-squares (best-fit) line? Interpret the slope.
  • Check student’s solution.
  • yes
  • ŷ = −266.8863+0.1656x
  • 0.9448; Yes
  • 62.8233; 62.3265
  • yes
  • no; (1987, 62.7)
  • 72.5937; no
  • slope = 0.1656.
    As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656.

11. The cost of a leading liquid laundry detergent in different sizes is given below.

Figure 9.32
Size (ounces) Cost ($) Cost per ounce
16 3.99
32 4.99
64 5.99
200 10.99
  1. Complete the table for the cost per ounce of the different sizes.
  2. Using “size” as the independent variable and “cost per ounce” as the dependent variable, draw a scatter plot of the data.
  3. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  4. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  5. Find the correlation coefficient. Is it significant?
  6. If the laundry detergent were sold in a 40-ounce size, find the estimated cost per ounce.
  7. If the laundry detergent were sold in a 90-ounce size, find the estimated cost per ounce.
  8. Does it appear that a line is the best way to fit the data? Why or why not?
  9. Are there any outliers in the the data?
  10. Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would cost per ounce? Why or why not?
  11. What is the slope of the least-squares (best-fit) line? Interpret the slope.
  • Figure 9.33
    Size (ounces) Cost ($) cents/oz
    16 3.99 24.94
    32 4.99 15.59
    64 5.99 9.36
    200 10.99 5.50
  • Check student’s solution.
  • There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size.
  • ŷ = 20.2368 – 0.0819x
  • r = –0.8086
  • 40-oz: 16.96 cents/oz
  • 90-oz: 12.87 cents/oz
  • The relationship is not linear; the least squares line is not appropriate.
  • no outliers
  • No, you would be extrapolating. The 300-oz size is outside the range of x.
  • slope = –0.08194; for each additional ounce in size, the cost per ounce decreases by 0.082 cents.

12. According to a flyer by a Prudential Insurance Company representative, the costs of approximate probate fees and taxes for selected net taxable estates are as follows:

Figure 9.34
Net Taxable Estate ($) Approximate Probate Fees and Taxes ($)
600,000 30,000
750,000 92,500
1,000,000 203,000
1,500,000 438,000
2,000,000 688,000
2,500,000 1,037,000
3,000,000 1,350,000
  1. Decide which variable should be the independent variable and which should be the dependent variable.
  2. Draw a scatter plot of the data.
  3. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  4. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx.
  5. Find the correlation coefficient. Is it significant?
  6. Find the estimated total cost for a next taxable estate of \$1,000,000. Find the cost for \$2,500,000.
  7. Does it appear that a line is the best way to fit the data? Why or why not?
  8. Are there any outliers in the data?
  9. Based on these results, what would be the probate fees and taxes for an estate that does not have any assets?
  10. What is the slope of the least-squares (best-fit) line? Interpret the slope.

13. The following are advertised sale prices of color televisions at Anderson’s.

Figure 9.35
Size (inches) Sale Price ($)
9 147
20 197
27 297
31 447
35 1177
40 2177
60 2497
  1. Decide which variable should be the independent variable and which should be the dependent variable.
  2. Draw a scatter plot of the data.
  3. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  4. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  5. Find the correlation coefficient. Is it significant?
  6. Find the estimated sale price for a 32 inch television. Find the cost for a 50 inch television.
  7. Does it appear that a line is the best way to fit the data? Why or why not?
  8. Are there any outliers in the data?
  9. What is the slope of the least-squares (best-fit) line? Interpret the slope.
  • Size is x, the independent variable, price is y, the dependent variable.
  • Check student’s solution.
  • The relationship does not appear to be linear.
  • ŷ = –745.252 + 54.75569x
  • r = 0.8944, yes it is significant
  • 32-inch: \$1006.93, 50-inch: \$1992.53
  • No, the relationship does not appear to be linear. However, r is significant.
  • no, the 60-inch TV
  • For each additional inch, the price increases by \$54.76

14. The figure below shows the average heights for American boys in 1990.

Figure 9.36
Age (years) Height (cm)
birth 50.8
2 83.8
3 91.4
5 106.6
7 119.3
10 137.1
14 157.5
  1. Decide which variable should be the independent variable and which should be the dependent variable.
  2. Draw a scatter plot of the data.
  3. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  4. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  5. Find the correlation coefficient. Is it significant?
  6. Find the estimated average height for a one-year-old. Find the estimated average height for an eleven-year-old.
  7. Does it appear that a line is the best way to fit the data? Why or why not?
  8. Are there any outliers in the data?
  9. Use the least squares line to estimate the average height for a sixty-two-year-old man. Do you think that your answer is reasonable? Why or why not?
  10. What is the slope of the least-squares (best-fit) line? Interpret the slope.

15. Use the table below to answer (a)-(n).

Figure 9.37
State # letters in name Year entered the Union Ranks for entering the Union Area (square miles)
Alabama 7 1819 22 52,423
Colorado 8 1876 38 104,100
Hawaii 6 1959 50 10,932
Iowa 4 1846 29 56,276
Maryland 8 1788 7 12,407
Missouri 8 1821 24 69,709
New Jersey 9 1787 3 8,722
Ohio 4 1803 17 44,828
South Carolina 13 1788 8 32,008
Utah 4 1896 45 84,904
Wisconsin 9 1848 30 65,499

We are interested in whether there is a relationship between the ranking of a state and the area of the state.

  1. What are the independent and dependent variables?
  2. What do you think the scatter plot will look like? Make a scatter plot of the data.
  3. Does it appear from inspection that there is a relationship between the variables? Why or why not?
  4. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx
  5. Find the correlation coefficient. What does it imply about the significance of the relationship?
  6. Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas?
  7. Use the two points in part f to plot the least-squares line on your graph from part b.
  8. Does it appear that a line is the best way to fit the data? Why or why not?
  9. Are there any outliers?
  10. Use the least squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not?
  11. Delete “Hawaii” and substitute “Alaska” for it. Alaska is the forty-ninth, state with an area of 656,424 square miles.
  12. Calculate the new least-squares line.
  13. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case?
  14. Do you think that, in general, newer states are larger than the original states?
  • Let rank be the independent variable and area be the dependent variable.
  • Check student’s solution.
  • There appears to be a linear relationship, with one outlier.
  • ŷ (area) = 24177.06 + 1010.478x
  • r = 0.50047, r is not significant so there is no relationship between the variables.
  • Alabama: 46407.576 Colorado: 62575.224
  • Alabama estimate is closer than Colorado estimate.
  • If the outlier is removed, there is a linear relationship.
  • There is one outlier (Hawaii).
  • rank 51: 75711.4; no
  • Figure 9.38
    Alabama 7 1819 22 52,423
    Colorado 8 1876 38 104,100
    Hawaii 6 1959 50 10,932
    Iowa 4 1846 29 56,276
    Maryland 8 1788 7 12,407
    Missouri 8 1821 24 69,709
    New Jersey 9 1787 3 8,722
    Ohio 4 1803 17 44,828
    South Carolina 13 1788 8 32,008
    Utah 4 1896 45 84,904
    Wisconsin 9 1848 30 65,499
  • ŷ = –87065.3 + 7828.532x
  • Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier.
  • yes, with the exception of Hawaii

3.5 Cautions About Regression

References

Image References

Figure 9.24: Figure 12.30 from OpenStax Introductory Statistics (2013) (CC BY 4.0). Retrieved from https://openstax.org/books/introductory-statistics/pages/12-practice

Text

Data from the House Ways and Means Committee, the Health and Human Services Department.

Data from Microsoft Bookshelf.

Data from the United States Department of Labor, the Bureau of Labor Statistics.

Data from the Physician’s Handbook, 1990.

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  2. United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013). ↵
  3. Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013). ↵
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  5. Samuel, T. M. “Strange Facts about RH Negative Blood.” Healthfully, 2017. Available online at https://healthfully.com/strange-rh-negative-blood-5552003.html (accessed January 26, 2021). ↵
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  11. Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). ↵
  12. Data from the San Jose Mercury News. ↵

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