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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)\operatorname { Pr } ( Z < 0.0 ) (b) Pr(Z1.0)\operatorname { Pr } ( Z \leq 1.0 ) (c) Pr(Z>1.96)\operatorname { Pr } ( Z > 1.96 ) (d) Pr(Z<2.0)\operatorname { Pr } ( Z < - 2.0 ) (e) Pr(Z>1.645)\operatorname { Pr } ( Z > 1.645 ) (f) Pr(Z>1.645)\operatorname { Pr } ( Z > - 1.645 ) (g) Pr(1.96<Z<1.96)\operatorname { Pr } ( - 1.96 < Z < 1.96 ) (h) Pr(Z<2.576\operatorname { Pr } ( Z < 2.576 or Z>2.576)Z > 2.576 ) (i) Pr(Z>z)=0.10\operatorname { Pr } ( Z > z ) = 0.10 ; find zz . (j) Pr(Z<z\operatorname { Pr } ( Z < - z or Z>z)=0.05Z > z ) = 0.05 ; find zz .

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(a)0.5000
(b)0.8413
...

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Assume that Y is normally distributed N(μ,σ2) N \left( \mu , \sigma ^ { 2 } \right) Moving from the mean (μ) ( \mu ) 1.96 standard deviations to the left and 1.96 standard deviations to the right, then the area under the normal p.d.f. is


A) 0.67
B) 0.05
C) 0.95
D) 0.33

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Two variables are uncorrelated in all of the cases below, with the exception of a. being independent. b. having a zero covariance. c. σXYσX2σY2\left| \sigma _ { X Y } \right| \leq \sqrt { \sigma _ { X } ^ { 2 } \sigma _ { Y } ^ { 2 } } . d. E(YX)=0E ( Y \mid X ) = 0 .

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The Economic Report of the President gives the following age distribution of the United States population for the year 2000:  United States Population By Age Group, 2000 Outcome (age  category)  Under 5 515161920242544456465 and  over  Percentage 0.060.160.060.070.300.220.13\begin{array}{l}\text { United States Population By Age Group, } 2000\\\begin{array} { | l | l | l | l | l | l | l | l | } \hline \begin{array} { l } \text { Outcome (age } \\\text { category) }\end{array} & \text { Under 5 } & 5 - 15 & 16 - 19 & 20 - 24 & 25 - 44 & 45 - 64 & \begin{array} { l } 65 \text { and } \\\text { over }\end{array} \\\hline \text { Percentage } & 0.06 & 0.16 & 0.06 & 0.07 & 0.30 & 0.22 & 0.13 \\\hline\end{array}\end{array} Imagine that every person was assigned a unique number between 1 and 275,372,000 (the total population in 2000).If you generated a random number, what would be the probability that you had drawn someone older than 65 or under 16? Treating the percentages as probabilities, write down the cumulative probability distribution.What is the probability of drawing someone who is 24 years or younger?

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In econometrics, we typically do not rely on exact or finite sample distributions because a. we have approximately an infinite number of observations (think of re-sampling). b. variables typically are normally distributed. c. the covariances of Yi,YjY _ { i } , Y _ { j } are typically not zero. d. asymptotic distributions can be counted on to provide good approximations to the exact sampling distribution (given the number of observations available in most cases).

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Let Y be a random variable.Then var(Y)equals a. E[(YμY)2]\sqrt { E \left[ \left( Y - \mu _ { Y } \right) ^ { 2 } \right] } . b. E[(YμY)]E \left[ \left| \left( Y - \mu _ { Y } \right) \right| \right] . c. E[(YμY)2]E \left[ \left( Y - \mu _ { Y } \right) ^ { 2 } \right] . d. E[(YμY)]E \left[ \left( Y - \mu _ { Y } \right) \right] .

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The conditional expectation of Y given X,E(YX=x) X , E ( Y \mid X = x ) is calculated as follows:


A) i=1kyiPr(X=xiY=y) \sum _ { i = 1 } ^ { k } y _ { i } \operatorname { Pr } \left( X = x _ { i } \mid Y = y \right)
B) E[E(YX) ]E [ E ( Y \mid X ) ]
C) i=1kyiPr(Y=yiX=x) \sum _ { i = 1 } ^ { k } y _ { i } \operatorname { Pr } \left( Y = y _ { i } \mid X = x \right)
D) i=1lE(YX=xi) Pr(X=xi) \sum _ { i = 1 } ^ { l } E \left( Y \mid X = x _ { i } \right) \operatorname { Pr } \left( X = x _ { i } \right)

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The variance of Yˉ,σYˉ2\bar { Y } , \sigma _ { \bar { Y } } ^ { 2 } is given by the following formula:


A) σY2\sigma _ { Y } ^ { 2 }
B) σYn\frac { \sigma _ { Y } } { \sqrt { n } }
C) σY2n\frac { \sigma _ { Y } ^ { 2 } } { n }
D) σY2n\frac { \sigma _ { Y } ^ { 2 } } { \sqrt { n } }

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The accompanying table gives the outcomes and probability distribution of the number of times a student checks her e-mail daily:  Probability of Checking E-Mail  Outcome  (number of e-  mail checks) 0123456 Probability  distribution 0.050.150.300.250.150.080.02\begin{array}{l}\text { Probability of Checking E-Mail }\\\begin{array} { | l | l | l | l | l | l | l | l | } \hline \begin{array} { l } \text { Outcome } \\\text { (number of e- } \\\text { mail checks) }\end{array} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \begin{array} { l } \text { Probability } \\\text { distribution }\end{array} & 0.05 & 0.15 & 0.30 & 0.25 & 0.15 & 0.08 & 0.02 \\\hline\end{array}\end{array} Sketch the probability distribution.Next, calculate the c.d.f.for the above table.What is the probability of her checking her e-mail between 1 and 3 times a day? Of checking it more than 3 times a day?

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The skewness is most likely positive for one of the following distributions:


A) The grade distribution at your college or university.
B) The U.S.income distribution.
C) SAT scores in English.
D) The height of 18-year-old females in the U.S.

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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=yX=x)=Pr(X=x,Y=y)Pr(X=x)\operatorname { Pr } ( Y = y \mid X = x ) = \frac { \operatorname { Pr } ( X = x , Y = y ) } { \operatorname { Pr } ( X = x ) } , derive a relationship between Pr(X=xY=y) and Pr(Y=yX=x)\operatorname { Pr } ( X = x \mid Y = y ) \text { and } \operatorname { Pr } ( Y = y \mid X = x ) This is called Bayes' theorem.

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You have read about the so-called catch-up theory by economic historians, whereby nations that are further behind in per capita income grow faster subsequently.If this is true systematically, then eventually laggards will reach the leader.To put the theory to the test, you collect data on relative (to the United States)per capita income for two years, 1960 and 1990, for 24 OECD countries.You think of these countries as a population you want to describe, rather than a sample from which you want to infer behavior of a larger population. The relevant data for this question is as follows:  You have read about the so-called catch-up theory by economic historians, whereby nations that are further behind in per capita income grow faster subsequently.If this is true systematically, then eventually laggards will reach the leader.To put the theory to the test, you collect data on relative (to the United States)per capita income for two years, 1960 and 1990, for 24 OECD countries.You think of these countries as a population you want to describe, rather than a sample from which you want to infer behavior of a larger population. The relevant data for this question is as follows:   where   X_{1}   and   X_{2}   are per capita income relative to the United States in 1960 and 1990 respectively, and   Y   is the average annual growth rate in   X   over the   1960-1990   period. Numbers in the last row represent sums of the columns above.  (a) Calculate the variance and standard deviation of   X_{1}   and   X_{2}  . For a catch-up effect to be present, what relationship must the two standard deviations show? Is this the case here? where X1 X_{1} and X2 X_{2} are per capita income relative to the United States in 1960 and 1990 respectively, and Y Y is the average annual growth rate in X X over the 19601990 1960-1990 period. Numbers in the last row represent sums of the columns above. (a) Calculate the variance and standard deviation of X1 X_{1} and X2 X_{2} . For a catch-up effect to be present, what relationship must the two standard deviations show? Is this the case here?

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The correlation coefficient is -0.88.It ...

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Probabilities and relative frequencies are related in that the probability of an outcome is the proportion of the time that the outcome occurs in the long run.Hence concepts of joint, marginal, and conditional probability distributions stem from related concepts of frequency distributions. You are interested in investigating the relationship between the age of heads of households and weekly earnings of households.The accompanying data gives the number of occurrences grouped by age and income.You collect data from 1,744 individuals and think of these individuals as a population that you want to describe, rather than a sample from which you want to infer behavior of a larger population.After sorting the data, you generate the accompanying table: Probabilities and relative frequencies are related in that the probability of an outcome is the proportion of the time that the outcome occurs in the long run.Hence concepts of joint, marginal, and conditional probability distributions stem from related concepts of frequency distributions. You are interested in investigating the relationship between the age of heads of households and weekly earnings of households.The accompanying data gives the number of occurrences grouped by age and income.You collect data from 1,744 individuals and think of these individuals as a population that you want to describe, rather than a sample from which you want to infer behavior of a larger population.After sorting the data, you generate the accompanying table:   The median of the income group of $800 and above is $1,050. (a)Calculate the joint relative frequencies and the marginal relative frequencies.Interpret one of each of these.Sketch the cumulative income distribution. The median of the income group of $800 and above is $1,050. (a)Calculate the joint relative frequencies and the marginal relative frequencies.Interpret one of each of these.Sketch the cumulative income distribution.

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The conditional distribution of Y given X=x,Pr(Y=yX=x) X = x , \operatorname { Pr } ( Y = y \mid X = x ) is


A) Pr(Y=y) Pr(X=x) \frac { \operatorname { Pr } ( Y = y ) } { \operatorname { Pr } ( X = x ) }
B) i=1lPr(X=xi,Y=y) \sum _ { i = 1 } ^ { l } \operatorname { Pr } \left( X = x _ { i } , Y = y \right)
C) Pr(X=x,Y=y) Pr(Y=y) \frac { \operatorname { Pr } ( X = x , Y = y ) } { \operatorname { Pr } ( Y = y ) }
D) Pr(X=x,Y=y) Pr(X=x) \frac { \operatorname { Pr } ( X = x , Y = y ) } { \operatorname { Pr } ( X = x ) }

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Following Alfred Nobel's will, there are five Nobel Prizes awarded each year.These are for outstanding achievements in Chemistry, Physics, Physiology or Medicine, Literature, and Peace.In 1968, the Bank of Sweden added a prize in Economic Sciences in memory of Alfred Nobel.You think of the data as describing a population, rather than a sample from which you want to infer behavior of a larger population.The accompanying table lists the joint probability distribution between recipients in economics and the other five prizes, and the citizenship of the recipients, based on the 1969-2001 period. Joint Distribution of Nobel Prize Winners in Economics and Non-Economics Disciplines, and Citizenship, 1969-2001  U.S. Citizen (Y=0) Non-U.S. Citizen (Y=1) Total  Economics Nobel  Prize (X=0)0.1180.0490.167 Physics, Chemistry,  Medicine, Literature,  and Peace Nobel  Prize (X=1)0.3450.4880.833 Total 0.4630.5371.00\begin{array}{|c|c|c|c|}\hline & \begin{array}{c}\text { U.S. Citizen } \\(Y=0)\end{array} & \begin{array}{c}\text { Non-U.S. Citizen } \\(Y=1)\end{array} & \text { Total } \\\hline \begin{array}{c}\text { Economics Nobel } \\\text { Prize }(X=0)\end{array} & 0.118 & 0.049 & 0.167 \\\hline \begin{array}{c}\text { Physics, Chemistry, } \\\text { Medicine, Literature, } \\\text { and Peace Nobel } \\\text { Prize }(X=1)\end{array} & 0.345 & 0.488 & 0.833 \\\hline \text { Total } &0.463 &0.537 &1.00 \\\hline\end{array} (a)Compute EY( )and interpret the resulting number.

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Two random variables are independently distributed if their joint distribution is the product of their marginal distributions.It is intuitively easier to understand that two random variables are independently distributed if all conditional distributions of Y given X are equal.Derive one of the two conditions from the other.

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If all conditional d...

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You are at a college of roughly 1,000 students and obtain data from the entire freshman class (250 students)on height and weight during orientation.You consider this to be a population that you want to describe, rather than a sample from which you want to infer general relationships in a larger population.Weight (Y)is measured in pounds and height (X)is measured in inches.You calculate the following sums: i=1nyi2=94,228.8,i=1nxi2=1,248.9,i=1nxiyi=7,625.9\sum _ { i = 1 } ^ { n } y _ { i } ^ { 2 } = 94,228.8 , \sum _ { i = 1 } ^ { n } x _ { i } ^ { 2 } = 1,248.9 , \sum _ { i = 1 } ^ { n } x _ { i } y _ { i } = 7,625.9  (small letters refer to deviations from means as in zi=ZiZˉ ). \text { (small letters refer to deviations from means as in } z _ { i } = Z _ { i } - \bar { Z } \text { ). } (a)Given your general knowledge about human height and weight of a given age, what can you say about the shape of the two distributions?

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In considering the purchase of a certain stock, you attach the following probabilities to possible changes in the stock price over the next year.  Stock Price Change During  Next Twelve Months (%)  Probability +150.2+50.300.450.05150.05\begin{array} { | l | l | } \hline \begin{array} { l } \text { Stock Price Change During } \\\text { Next Twelve Months (\%) }\end{array} & \text { Probability } \\\hline + 15 & 0.2 \\\hline + 5 & 0.3 \\\hline 0 & 0.4 \\\hline - 5 & 0.05 \\\hline - 15 & 0.05 \\\hline\end{array} What is the expected value, the variance, and the standard deviation? Which is the most likely outcome? Sketch the cumulative distribution function. .

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The expectations augmented Phillips curve postulates Δp=πf(uuˉ),\Delta p = \pi - f ( u - \bar { u } ) , where Δp\Delta p is the actual inflation rate, π\pi is the expected inflation rate, and uu is the unemployment rate, with "-" indicating equilibrium (the NAIRU - Non-Accelerating Inflation Rate of Unemployment). Under the assumption of static expectations (π=Δp1)\left( \pi = \Delta p _ { - 1 } \right) , i.e., that you expect this period's inflation rate to hold for the next period ("the sun shines today, it will shine tomorrow"), then the prediction is that inflation will accelerate if the unemployment rate is below its equilibrium level. The accompanying table below displays information on accelerating annual inflation and unemployment rate differences from the equilibrium rate (cyclical unemployment), where the latter is approximated by a five-year moving average. You think of this data as a population which you want to describe, rather than a sample from which you want to infer behavior of a larger population. The data is collected from United States quarterly data for the period 1964:11964 : 1 to 1995:41995 : 4 . Joint Distribution of Accelerating Inflation and Cyclical Unemployment, 1964:1-1995:4 (uuˉ)>0(uuˉ)0 Total (Y=0)(Y=1)ΔpΔp1>00.1560.3830.539(X=0)ΔpΔp100.2970.1640.461(X=1) Total 0.4530.5471.00\begin{array}{|c|c|c|c|} \hline& (u-\bar{u})>0 & (u-\bar{u}) \geq 0 & \text { Total } \\& (Y=0) & (Y=1) & \\\hline \Delta p-\Delta p_{-1}>0 & 0.156 & 0.383 & 0.539 \\(X=0) & & & \\\hline \Delta p-\Delta p_{-1} \leq 0 & 0.297 & 0.164 & 0.461 \\(X=1) & & & \\\hline \text { Total } & 0.453 & 0.547 & 1.00 \\\hline\end{array} (a)Compute EY( )and EX( ), and interpret both numbers.

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There is a 65.6 percent probab...

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You consider visiting Montreal during the break between terms in January.You go to the relevant Web site of the official tourist office to figure out the type of clothes you should take on the trip.The site lists that the average high during January is -70 C, with a standard deviation of 40 C.Unfortunately you are more familiar with Fahrenheit than with Celsius, but find that the two are related by the following linear function: C=59(F32).C = \frac { 5 } { 9 } ( F - 32 ) . Find the mean and standard deviation for the January temperature in Montreal in Fahrenheit.

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Using equations (2.2...

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