Exam 2: Review of Probability

arrow
  • Select Tags
search iconSearch Question
  • Select Tags

The table accompanying lists the joint distribution of unemployment in the United States in 2001 by demographic characteristics (race and gender). Joint Distribution of Unemployment by Demographic Characteristics, United States, 2001 White (Y=0) Black and Other (Y=1) Total Age 16-19 (X=0) 0.13 0.05 0.18 Age 20 and above (X=1) 0.60 0.22 0.82 Total 0.73 0.27 1.00 (a)What is the percentage of unemployed white teenagers? (b)Calculate the conditional distribution for the categories "white" and "black and other." (c)Given your answer in the previous question, how do you reconcile this fact with the probability to be 60% of finding an unemployed adult white person, and only 22% for the category "black and other."

Free
(Essay)
5.0/5
(30)
Correct Answer:
Verified

(a)Pr(Y = 0, X = 0)= 0.13.
(b)
Conditional Distribution of Unemployment by Demographic
Characteristics, United States, 2001  White (Y=0) Black and Other (Y=1) Age 16-19 (X=0)0.180.19 Age 20 and above (X=1)0.820.81 Total 1.001.00\begin{array} { | c | c | c | } \hline & \begin{array} { c } \text { White } \\( Y = 0 )\end{array} & \begin{array} { c } \text { Black and Other } \\( Y = 1 )\end{array} \\\hline \begin{array} { c } \text { Age 16-19 } \\( X = 0 )\end{array} & 0.18 & 0.19 \\\hline \begin{array} { c } \text { Age 20 and above } \\( X = 1 )\end{array} & 0.82 & 0.81 \\\hline \text { Total } & 1.00 & 1.00 \\\hline\end{array} (c)The original table showed the joint probability distribution, while the table in (b)presented the conditional probability distribution.

Explain why the two probabilities are identical for the standard normal distribution: Pr(-1.96 ≤ X ≤ 1.96)and Pr(-1.96 < X < 1.96).

Free
(Essay)
4.8/5
(29)
Correct Answer:
Verified

For a continuous distribution, the probability of a point is zero.

There are frequently situations where you have information on the conditional distribution of Y given X, but are interested in the conditional distribution of X given Y. Recalling Pr(Y = y X\mid X = x)= Pr(X=x,Y=y)Pr(X=x)\frac { \operatorname { Pr } ( X = x , Y = y ) } { \operatorname { Pr } ( X = x ) } , derive a relationship between Pr(X = x Y\mid Y = y)and Pr(Y = y X\mid X = x). This is called Bayes' theorem.

Free
(Essay)
4.9/5
(35)
Correct Answer:
Verified

Given Pr(Y = y X\mid X = x)= Pr(X=xY=y)Pr(X=x)\frac { \operatorname { Pr } ( X = x \mid Y = y ) } { \operatorname { Pr } ( X = x ) } ,
Pr(Y = y X\mid X = x)× Pr(X = x)= Pr(X = x, Y = y);
similarly Pr(X = x Y\mid Y = y)= Pr(X=xY=y)Pr(X=x)\frac { \operatorname { Pr } ( X = x \mid Y = y ) } { \operatorname { Pr } ( X = x ) } and
Pr(X = x Y\mid Y = y)× Pr(Y = y)= Pr(X = x, Y = y). Equating the two and solving for Pr(X = x Y\mid Y = y)then results in
Pr(X = x Y\mid Y = y)= Pr(Y=yX=x)×Pr(X=x)Pr(Y=y)\frac { \operatorname { Pr } ( Y = y \mid X = x ) \times \operatorname { Pr } ( X = x ) } { \operatorname { Pr } ( Y = y ) } .

What is the probability of the following outcomes? (a)Pr(M = 7) (b)Pr(M = 2 or M = 10) (c)Pr(M = 4 or M ≠ 4) (d)Pr(M = 6 and M = 9) (e)Pr(M < 8) (f)Pr(M = 6 or M > 10)

(Essay)
4.8/5
(32)

Calculate the following probabilities using the standard normal distribution. Sketch the probability distribution in each case, shading in the area of the calculated probability. (a)Pr(Z < 0.0) (b)Pr(Z ≤ 1.0) (c)Pr(Z > 1.96) (d)Pr(Z < -2.0) (e)Pr(Z > 1.645) (f)Pr(Z > -1.645) (g)Pr(-1.96 < Z < 1.96) (h.)Pr(Z < 2.576 or Z > 2.576) (i.)Pr(Z > z)= 0.10; find z. (j.)Pr(Z < -z or Z > z)= 0.05; find z.

(Essay)
4.8/5
(41)

The expected value of a discrete random variable

(Multiple Choice)
4.8/5
(39)

Find the following probabilities: (a)Y is distributed X42X _ { 4 } ^ { 2 } Find Pr(Y > 9.49). (b)Y is distributed t. Find Pr(Y > -0.5). (c)Y is distributed F4,. Find Pr(Y < 3.32). (d)Y is distributed N(500, 10000). Find Pr(Y > 696 or Y < 304).

(Essay)
4.8/5
(37)

The correlation between X and Y

(Multiple Choice)
4.9/5
(39)

The sample average is a random variable and

(Multiple Choice)
4.8/5
(31)

Use the definition for the conditional distribution of Y given X = x and the marginal distribution of X to derive the formula for Pr(X = x, Y = y). This is called the multiplication rule. Use it to derive the probability for drawing two aces randomly from a deck of cards (no joker), where you do not replace the card after the first draw. Next, generalizing the multiplication rule and assuming independence, find the probability of having four girls in a family with four children.

(Essay)
4.9/5
(38)

Use a standard spreadsheet program, such as Excel, to find the following probabilities from various distributions analyzed in the current chapter: a. If Y is distributed N (1,4), find Pr(Y ≤ 3) b. If Y is distributed N (3,9), find Pr(Y > 0) c. If Y is distributed N (50,25), find Pr(40 ≤ Y ≤ 52) d. If Y is distributed N (5,2), find Pr(6 ≤ Y ≤ 8)

(Essay)
4.8/5
(32)

The variance of Yˉ,σ2Y\bar { Y } , \sigma \frac { 2 } { Y } , is given by the following formula:

(Multiple Choice)
5.0/5
(33)

The Student t distribution is

(Multiple Choice)
4.8/5
(42)

Looking at a large CPS data set with over 60,000 observations for the United States and the year 2004, you find that the average number of years of education is approximately 13.6. However, a surprising large number of individuals (approximately 800)have quite a low value for this variable, namely 6 years or less. You decide to drop these observations, since none of your relatives or friends have that few years of education. In addition, you are concerned that if these individuals cannot report the years of education correctly, then the observations on other variables, such as average hourly earnings, can also not be trusted. As a matter of fact you have found several of these to be below minimum wages in your state. Discuss if dropping the observations is reasonable.

(Essay)
4.8/5
(33)

The cumulative probability distribution shows the probability

(Multiple Choice)
4.9/5
(35)

var(aX + bY)=

(Multiple Choice)
4.7/5
(22)

The following problem is frequently encountered in the case of a rare disease, say AIDS, when determining the probability of actually having the disease after testing positively for HIV. (This is often known as the accuracy of the test given that you have the disease.)Let us set up the problem as follows: Y = 0 if you tested negative using the ELISA test for HIV, Y = 1 if you tested positive; X = 1 if you have HIV, X = 0 if you do not have HIV. Assume that 0.1 percent of the population has HIV and that the accuracy of the test is 0.95 in both cases of (i)testing positive when you have HIV, and (ii)testing negative when you do not have HIV. (The actual ELISA test is actually 99.7 percent accurate when you have HIV, and 98.5 percent accurate when you do not have HIV.) (a)Assuming arbitrarily a population of 10,000,000 people, use the accompanying table to first enter the column totals. Test Positive (Y=1) Test Negative (Y=0) Total HIV (X=1) No HIV (X=0) Total 10,000,000 (b)Use the conditional probabilities to fill in the joint absolute frequencies. (c)Fill in the marginal absolute frequencies for testing positive and negative. Determine the conditional probability of having HIV when you have tested positive. Explain this surprising result. (d)The previous problem is an application of Bayes' theorem, which converts Pr(Y = y X\mid X = x)into Pr(X = x Y\mid Y = y). Can you think of other examples where Pr(Y = y X\mid X = x)? Pr(X = x Y\mid Y = y)?

(Essay)
4.8/5
(36)

When there are ? degrees of freedom, the t? distribution

(Multiple Choice)
4.8/5
(41)

The accompanying table lists the outcomes and the cumulative probability distribution for a student renting videos during the week while on campus. Video Rentals per Week during Semester Outcome (number of weekly video rentals) 0 1 2 3 4 5 Probability distribution 0.05 0.55 0.25 0.05 0.07 0.02 Sketch the probability distribution. Next, calculate the cumulative probability distribution for the above table. What is the probability of the student renting between 2 and 4 a week? Of less than 3 a week?

(Essay)
4.8/5
(35)

The textbook mentioned that the mean of Y, E(Y)is called the first moment of Y, and that the expected value of the square of Y, E(Y2)is called the second moment of Y, and so on. These are also referred to as moments about the origin. A related concept is moments about the mean, which are defined as E[(Y -µY)r]. What do you call the second moment about the mean? What do you think the third moment, referred to as "skewness," measures? Do you believe that it would be positive or negative for an earnings distribution? What measure of the third moment around the mean do you get for a normal distribution?

(Essay)
4.7/5
(32)
Showing 1 - 20 of 70
close modal

Filters

  • Essay(0)
  • Multiple Choice(0)
  • Short Answer(0)
  • True False(0)
  • Matching(0)