1280 self Quiz UNIT 4 - these are self quizzes and are helpful when reviewing for the exam or graded PDF

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these are self quizzes and are helpful when reviewing for the exam or graded quizzes...

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UNIT 4 In Chapter 3 the data frame "ex.2", which contained information associated to the blood pressure of a sample of 150 men and women, was introduced. Let us assume that this sample was taken from an imaginary population of size 100,000 and let the information for all the members of this population be stored in a CSV file by the name "pop2.csv". Read the content of the population file into an data frame under the name "pop.2". Applying the function "summary" to this data frame produces: > summary(pop.2) id

sex

age

Min. : 1000050

FEMALE:50032

Min. :20.00

1st Qu.: 3227517

MALE :49968

1st Qu.:32.00

Median : 5479268

Median :35.00

Mean : 5482739

Mean :34.98

3rd Qu.: 7721879

3rd Qu.:38.00

Max. : 9999889

Max. :54.00

bmi

systolic

diastolic

Min. : 9.986

Min. : 73.37

Min. : 24.77

1st Qu.:22.081

1st Qu.:116.33

1st Qu.: 72.62

Median :24.819

Median :124.64

Median : 81.27

Mean :24.984

Mean :125.02

Mean : 81.67

3rd Qu.:27.704

3rd Qu.:133.22

3rd Qu.: 90.30

Max. :46.232

Max. :191.65

Max. :152.34

group HIGH :28126 LOW : 4215 NORMAL:67659 The variables in this data frame are the same variables that were included in the data frame "ex.2" of Chapter 3. The variables that are included in this data frame are: id:

A numerical variable. A 7 digits number that serves as a unique identifier of the subject. sex: A factor variable. The sex of each subject. The values are either "MALE" or "FEMALE". age: A numerical variable. The age of each subject. bmi: A numerical variable. The body mass index of each subject. systolic: A numerical variable. The systolic blood pressure of each subject. diastolic: A numerical variable. The diastolic blood pressure of each subject. group: A factor variable. The blood pressure category of each subject. The values are "NORMAL" both the systolic blood pressure is within its normal range (between 90 and 139) and the diastolic blood pressure is within its normal range (between 60 and 89). The value is "HIGH" if either measurements of blood pressure are above their normal upper limits and it is "LOW" if either measurements are below their normal lower limits. The next 6 questions correspond to a person that is sampled at random from this population. The answers below may be rounded up to two decimal places.

Question 1 Correct Mark 1.00 out of 1.00 Question text The median age in this sample was: Answer: 35

Feedback Run the code: > pop.2 median(pop.2$age) [1] 35

Question 2 Correct Question text The variance of the diastolic blood pressure in this sample was: Answer: 171.6469

Feedback Run the code: > pop.2 var(pop.2$diastolic) [1] 171.6469 Question 3 Correct Mark 1.00 out of 1.00 Question text The standard deviation of the difference between the 3.9507

systolic and diastolic blood pressures was: Answer (Hint: Observe that the difference " pop.2$systolic -

pop.2$diastolic " produces the difference between the two types of blood pressure for all the members in the population.) Feedback Run the code: > pop.2 sd(pop.2$systolic - pop.2$diastolic) [1] 3.950757 Question 4 Correct Mark 1.00 out of 1.00 Question text The largest BMI value in this sample was: Answer: 46.23199

Feedback You may ru the code: > pop.2 summary(pop.2$bmi) Min. 1st Qu. Median Mean 3rd Qu. Max. 9.986 22.080 24.820 24.980 27.700 46.230 Or you may use > max(pop.2$bmi) [1] 46.23199 Question 5 Correct Mark 1.00 out of 1.00 Question text The probability that someone sampled from this data will have normal blood pressure is: Answer: 0.67659

Feedback Run the code: > pop.2 summary(pop.2$group) HIGH LOW NORMAL 28126 4215 67659 There are 67,659 individuals that are classified as "NORMAL" among the total population of 100,000. Hence, the probability is 0.67659. Question 6 Correct Mark 1.00 out of 1.00 Question text Mark the following statement as either TRUE or FALSE: The standard deviation of the difference between the systolic and diastolic blood pressures is equal to the difference between the standard deviation of systolic blood pressure and the standard deviation of diastolic blood pressure Answer

Feedback

False

.

Run the code: > pop.2 sd(pop.2$systolic - pop.2$diastolic) [1] 3.950757 > sd(pop.2$systolic) - sd(pop.2$diastolic) [1] -0.6111588 Information Information text

The Distribution of Y Value 1.5 4 5.5 6 7.5 10

Probability 0.15 0.10 0.23 0.11 0.05 Complete the probabilities of the random variable Y in above table and compute

Question 7 Correct Mark 1.00 out of 1.00 Question text P(Y < 5) equals Answer: 0.51

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 The event in question involves the values 1.5 and 4. Hence P(Y < 5) = 0.15 + 0.36 = 0.51. Question 8 Correct Mark 1.00 out of 1.00 Question text P(Y not an integer) equals

Answer: 0.36

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 The event in question involves the values 1.5 and 5.5 and 7.5. Hence P(not an integer) = 0.15 + 0.10 + 0.11 = 0.36. Question 9 Correct Mark 1.00 out of 1.00 Question text E(Y) equals Answer: 4.92

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 Run the code: > Y.val P.val sum(Y.val*P.val) [1] 4.92 Question 10 Correct Mark 1.00 out of 1.00 Question text The standard deviation of Y is equal to Answer: 2.093705

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 Run the code: > Y.val P.val E Var sqrt(Var) [1] 2.093705 Information Information text One selects an integer between 1 and 9 (including 1 and 9) at random. Let X be a random variable that obtains as a value the integer that was selected. The following questions correspond to this random variable.

The Distribution of X Value

Probability

Complete table of distribution of X and compute: Question 11 Correct Mark 1.00 out of 1.00 Question text E(X) is equal to Answer: 5

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In order to compute the expectation run the code: > X.val P.val sum(X.val*P.val) [1] 5 Question 12 Correct Mark 1.00 out of 1.00 Question text The variance of X is equal to Answer: 6.666667

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In order to compute the variance run the code: > X.val P.val E sum((X.val-E)^2*P.val) [1] 6.666667 Question 13 Correct Mark 1.00 out of 1.00 Question text The standard deviation of X is equal to Answer: 2.581989

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In order to compute the variance run the code: > X.val P.val E Var sqrt(Var) [1] 2.581989 Information Information text

Suppose that you are offered the following "deal." An impartial person selects an integer between 1 and 9 at random and you try to guess beforehand which number will be selected. If you guess wrong then you pay $1 and if you guess right then you win $10. Call the outcome of the game your "gain." (Note that if you pay money then your gain is negative.) Question 14 Incorrect Mark 0.00 out of 1.00 Question text In this game your probability of winning is Select one: a. Larger than your probability of losing. b. Smaller than your probability of losing. c. Equal to your probability of losing.

d. Unknown. Feedback You select a number. The probability that the specific number that you have selected will turn out is 1/9. The probability that you miss is 8/9. Question 15 Correct Mark 1.00 out of 1.00 Question text The expected gain in this game is Answer: 0.2222

Feedback You select a number. The probability that the specific number that you have selected will turn out is 1/9. The probability that you miss is 8/9. Let X be the gain from the gain. The gain is 10 if you win (with probability 1/9) and -1

if you loose (with probability 8/9). The expectation is E(X) = 10/9 + (-1)*8/9 = 2/9 = 0.2222 UNIT 4 In Chapter 3 the data frame "ex.2", which contained information associated to the blood pressure of a sample of 150 men and women, was introduced. Let us assume that this sample was taken from an imaginary population of size 100,000 and let the information for all the members of this population be stored in a CSV file by the name "pop2.csv". Read the content of the population file into an data frame under the name "pop.2". Applying the function "summary" to this data frame produces: > summary(pop.2) id

sex

age

Min. : 1000050

FEMALE:50032

Min. :20.00

1st Qu.: 3227517

MALE :49968

1st Qu.:32.00

Median : 5479268

Median :35.00

Mean : 5482739

Mean :34.98

3rd Qu.: 7721879

3rd Qu.:38.00

Max. : 9999889

Max. :54.00

bmi

systolic

diastolic

Min. : 9.986

Min. : 73.37

Min. : 24.77

1st Qu.:22.081

1st Qu.:116.33

1st Qu.: 72.62

Median :24.819

Median :124.64

Median : 81.27

Mean :24.984

Mean :125.02

Mean : 81.67

3rd Qu.:27.704

3rd Qu.:133.22

3rd Qu.: 90.30

Max. :46.232

Max. :191.65

Max. :152.34

group HIGH :28126 LOW : 4215 NORMAL:67659

The variables in this data frame are the same variables that were included in the data frame "ex.2" of Chapter 3. The variables that are included in this data frame are: id: A numerical variable. A 7 digits number that serves as a unique identifier of the subject. sex: A factor variable. The sex of each subject. The values are either "MALE" or "FEMALE". age: A numerical variable. The age of each subject. bmi: A numerical variable. The body mass index of each subject. systolic: A numerical variable. The systolic blood pressure of each subject. diastolic: A numerical variable. The diastolic blood pressure of each subject. group: A factor variable. The blood pressure category of each subject. The values are "NORMAL" both the systolic blood pressure is within its normal range (between 90 and 139) and the diastolic blood pressure is within its normal range (between 60 and 89). The value is "HIGH" if either measurements of blood pressure are above their normal upper limits and it is "LOW" if either measurements are below their normal lower limits. The next 6 questions correspond to a person that is sampled at random from this population. The answers below may be rounded up to two decimal places.

Question 1 Correct Mark 1.00 out of 1.00 Question text The median age in this sample was: Answer: 35

Feedback Run the code: > pop.2 median(pop.2$age) [1] 35 Question 2 Correct Question text The variance of the diastolic blood pressure in this sample was: Answer: 171.6469

Feedback Run the code: > pop.2 var(pop.2$diastolic) [1] 171.6469 Question 3 Correct Mark 1.00 out of 1.00 Question text The standard deviation of the difference between the 3.9507

systolic and diastolic blood pressures was: Answer (Hint: Observe that the difference " pop.2$systolic -

pop.2$diastolic " produces the difference between the two types of blood pressure for all the members in the population.) Feedback Run the code: > pop.2 sd(pop.2$systolic - pop.2$diastolic) [1] 3.950757 Question 4 Correct Mark 1.00 out of 1.00 Question text The largest BMI value in this sample was:

Answer: 46.23199

Feedback You may ru the code: > pop.2 summary(pop.2$bmi) Min. 1st Qu. Median Mean 3rd Qu. Max. 9.986 22.080 24.820 24.980 27.700 46.230 Or you may use > max(pop.2$bmi) [1] 46.23199 Question 5 Correct Mark 1.00 out of 1.00 Question text The probability that someone sampled from this data will have normal blood pressure is: Answer: 0.67659

Feedback Run the code: > pop.2 summary(pop.2$group) HIGH LOW NORMAL 28126 4215 67659 There are 67,659 individuals that are classified as "NORMAL" among the total population of 100,000. Hence, the probability is 0.67659. Question 6 Correct Mark 1.00 out of 1.00 Question text Mark the following statement as either TRUE or FALSE: The standard deviation of the difference between the systolic and diastolic blood pressures is equal to the difference between the standard deviation of systolic blood pressure and the standard deviation of diastolic blood pressure Answer

False

.

Feedback Run the code: > pop.2 sd(pop.2$systolic - pop.2$diastolic) [1] 3.950757 > sd(pop.2$systolic) - sd(pop.2$diastolic) [1] -0.6111588 Information Information text

The Distribution of Y Value 1.5 4 5.5 6 7.5 10

Probability 0.15 0.10 0.23 0.11 0.05 Complete the probabilities of the random variable Y in above table and compute

Question 7 Correct Mark 1.00 out of 1.00 Question text P(Y < 5) equals Answer: 0.51

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 The event in question involves the values 1.5 and 4. Hence P(Y < 5) = 0.15 + 0.36 = 0.51. Question 8 Correct Mark 1.00 out of 1.00 Question text

P(Y not an integer) equals Answer: 0.36

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 The event in question involves the values 1.5 and 5.5 and 7.5. Hence P(not an integer) = 0.15 + 0.10 + 0.11 = 0.36. Question 9 Correct Mark 1.00 out of 1.00 Question text E(Y) equals Answer: 4.92

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 Run the code: > Y.val P.val sum(Y.val*P.val) [1] 4.92 Question 10 Correct Mark 1.00 out of 1.00 Question text The standard deviation of Y is equal to Answer: 2.093705

Feedback The missing probability is equal to 1 -(0.15 + 0.1 + 0.23 + 0.11 + 0.05) = 0.36 Run the code: > Y.val P.val E Var sqrt(Var) [1] 2.093705 Information Information text One selects an integer between 1 and 9 (including 1 and 9) at random. Let X be a random variable that obtains as a value the integer that was selected. The following questions correspond to this random variable.

The Distribution of X Value

Probability

Complete table of distribution of X and compute: Question 11 Correct Mark 1.00 out of 1.00 Question text E(X) is equal to Answer: 5

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In

order to compute the expectation run the code: > X.val P.val sum(X.val*P.val) [1] 5 Question 12 Correct Mark 1.00 out of 1.00 Question text The variance of X is equal to Answer: 6.666667

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In order to compute the variance run the code: > X.val P.val E sum((X.val-E)^2*P.val) [1] 6.666667 Question 13 Correct Mark 1.00 out of 1.00 Question text The standard deviation of X is equal to Answer: 2.581989

Feedback The values of the random variables are the integers between 1 and 9 and the probability of each value is 1/9 In order to compute the variance run the code: > X.val P.val E Var sqrt(Var) [1] 2.581989 Information

Information text Suppose that you are offered the following "deal." An impartial person selects an integer between 1 and 9 at random and you try to guess beforehand which number will be selected. If you guess wrong then you pay $1 and if you guess right then you win $10. Call the outcome of the game your "gain." (Note that if you pay money then your gain is negative.) Question 14 Incorrect Mark 0.00 out of 1.00 Question text In this game your probability of winning is Select one: a. Larger than your probability of losing. b. Smaller than your probability of losing. c. Equal to your probability of losing.

d. Unknown. Feedback You select a number. The probability that the specific number that you have selected will turn out is 1/9. The probability that you miss is 8/9. Question 15 Correct Mark 1.00 out of 1.00 Question text The expected gain in this game is Answer: 0.2222

Feedback You select a number. The probability that the specific number that you have selected will turn out is 1/9. The probability that you miss is 8/9. Let X be the gain from the

gain. The gain is 10 if you win (with probability 1/9) and -1 if you loose (with probability 8/9). The expectation is E(X) = 10/9 + (-1)*8/9 = 2/9 = 0.2222...