251y0032 11/28/00 ECO251 QBA1 Name _______key_________ THIRD EXAM Section Enrolled: MWF TR 10 11 12:30 NOVEMBER 21, 2000 Part I. Do all the Following (10+ Points) Show your work! Please indicate clearly what sections of the problem you are answering! If you are following a rule like xaEaxE please state it! If you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! Assume that the following joint probability table corresponds to two variables x and y 25.35.40.15.05.05.05.20.10.05.10.25.420210yyPxa. Are x and yindependent? Why?(2)b. Compute x, the standard deviation of x. (2)c. Compute xy, the covariance of x and y, and interpret it. (3)d. Compute xy, the correlation of x and y, and interpret it. (2)e. Find 2 yxP (2)f. yx can take the following values: 0, 1, 2, 3, 4, 5 and 6. Write out the distribution of yx by making a list with these values and their probabilities (2)g. (i) Find the mean and standard deviation of yx from the table you just made. (2) (ii) Show that they are the same as the values of the mean and standard deviation of yx that you would have gotten using the method you used in the last part of Graded Assignment 2. (2)Solution:a) Check for independence: First you need to find xP and yP. Look at the upper left hand probability below. Its value is .25 and it represents 00 yxP . If x and y were independent , we would have 00 yxP 16.40.40.00 yPxP. Since this is not true, we can say that x and y are not independent.. 40.570.135.185.00.100.135.000.050.035.000.025.35.40.00.440.100.000.170.000.025.35.40.15.05.05.05.20.10.05.10.25.42021022xPxxxPxPyyPyyyPyPx To summarize 1xP, 85.0xxPxEx, 35.122xPxxE, 1yP, 7.1yyPyEy and 4.522yPyyE251y0032 11/28/00 0.22.12.00.02.04.00.00.00.00.04215.4105.4005.2205.2120.2010.0205.0110.0025.xyxyPxyE 555.07.185.00.2 yxxyxyExyCov, 6275.085.035.12222xxxE and 51.27.14.52222yyyE. So that 44223.051.26275.0555.0yxxyxy. (79215.0x,5843.1y)b) Compute x, the standard deviation of x. (2) Above, it says 6275.085.035.12222xxxE and 79215.0xc) Compute xy, the covariance of x and y, and interpret it. (3) Above, it says 555.07.185.00.2 yxxyxyExyCov Since this is positive x and y seem to move together; if one rises, the other also tends to rise.d) Compute xy, the correlation of x and y, and interpret it. (2) Above, it says44223.051.26275.0555.0yxxyxy. If we square this correlation, we get .1956. Though the positive sign indicates that x and y seem to move together, .1956 is quite small on a zero to one scale, so the correlation is rather weak.251y0032 11/28/00e) Find 2 yxP (1) the original table is copied below, along with a second table with the sums ofx and y .15.05.05.05.20.10.05.10.25.420210yx 654432210420210yxThere are two ways to get 2: by adding 0x and 2y, or by adding 2x and 0y. The joint probability of the first pair is .10 and the joint probability of the second pair is .05, so the total probability is 15.05.10.2 yxPf), g) yx can take the following values: 0, 1, 2, 3, 4, 5 and 6. Write out the distribution of yx by making a list with these values and their probabilities (2) . (i) Find the mean and standard deviation ofyx from the table you just made. (2) (ii) Show that they are the same as the values of the mean and standard deviation of yx that you would have gotten using the method you used in the last part of Graded Assignment 2. (2) We put the probabilities together for each value of yx in the same way we did 4 yxP. The table follows: 75.1040.525.160.180.160.10.055.200.190.15.625.05.540.10.460.20.330.15.210.10.1025.02yxPyxtotalyxPyxyxPyx So 55.2 yxE and 75.102 yxE. This means that yxVar 2475.455.275.10222 yxEyxEand 0609.22475.4 yx. If we compute yxE and yxVar from the results of b-d we get exactly the same results as we did above. 55.27.185.0 yxyExEyxE and xyyxyxVar222 yxCovyVarxVar ,2 2475.4555.0251.26275.0 .4251y0032 11/28/00Part II. Do the following problems. ( Do at least 40 points – but remember anything beyond 40 points is extra credit. ). Show your work! Please indicate clearly what sections of a problem you are answering! And what columns you found table values in.. If you are following a rule like xaEaxE please state it! If you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! You must do part a) of problem 3! (3 point penalty!)1. Describe the meaning and give the probabilities for the problems below.Example: For the Hypergeometric Distribution, P x1 2 when N n M 18 5 9, , and is the probability of between 1and 2 successes when a sample of 5 is taken from a population of 18 where there are 9 successes in the population. !13!5!18!6!3!9!7!2!9!13!5!18!5!4!9!8!1!92118593921859491CCCCCCxP 48529.a. Binomial 104 xP, when 35.p and 20n(2)b. Binomial 104 xP, when 85.p and 20n(2)c. Poisson 104 xP, when 8m (2)d. Geometric 104 xP, when 35.p (2)e. No meaning is needed for the
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