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WCU ECO 251 - ECO 251 Final Exam

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251y0341 05/05/03 ECO 251 QBA1 Name KEY FINAL EXAM Class ________________ MAY 7, 2003Part I. Do all the Following (14 Points) Make Diagrams! Show your work! 4,3~ Nx. How many of you really believe that a probability can be negative? 1.  85.130.2  xP 29.032.14385.14330.2 zPzP   029.0032.1  zPzP2925.1141.4066. or    029.0033.1  zPzP2941.1141.4082.. or , better,    2933.1141.4074.029.00325.1  zPzP 2.  171 xP 50.350.04317431 zPzP   6913.4998.1915.50.30050.0  zPzP3.  85.1xP     0029.029.04385.1 zPzPzPzP 6141.5.1141.  4.  00.4F (Cumulative Probability)  4344 zPxP     5987.0987.5.25.00025.0  zPzPzP5.  00.400.4  xP 25.075.1434434 zPzP   5586.0987.4599.25.00075.1  zPzP6. 015.x (Find 015.z first) Make a diagram for z. Show a Normal curve with a mean of zero in its center. Remember that 015.z is a point with 1.5% above it and 98.5% below it. Since 50% of the distribution is bellow zero  .4850.5.9850.0015. zzPAccording to the Normal table 4850.17.20 zP. So 17.2015.z and 68.1168.83417.23015.zx. Check:  68.11xP     0150.4850.5.17.20017.24368.11 zPzPzPzP7. A symmetrical region around the mean with a probability of 25%. Make a diagram for z. Show a Normal curve with a mean of zero in its center. If we split 25% in two, we get two areas, one on either side of the mean with probabilities of 12.5%. We can call the point we want 375.z, because, since the area above zero is 50%, the area above 375.z must be 50% - 12.5% = 37.5%. But we have already decided that the probability between 375.z and zero is 12.5%. The closest we can come on the Normal table is  1255.32.00 zP, so 32.0375.z and  28.13432.03375.zxor 1.72 to 4.28,1Check:       1255.232.00232.032.04328.44372.128.472.1  zPzPzPxP2510.Exam is normed on 75 points. There are actually 128 possible points.2251y0341 05/05/03 II. (10 points+-2 point penalty for not trying part a .) Show your work! The following numbers apply to 9 developed countries and give deaths per 100 million miles and speed limits. (xy was not supplied and is calculated in red.) Row deaths SpLim x y xy 1 3.1 55 170.5 2 3.4 55 187.0 3 3.5 55 192.5 4 3.6 70 252.0 5 4.2 55 231.0 6 4.4 60 264.0 7 4.8 55 264.0 8 5.0 60 300.0 9 6.2 75 465.02326.0These sums have been calculated for you. 2.38x, 86.1692x, 540yand328502y. Please calculate the following:a. The sample standard deviation of x (4) Note that 00.60y and 50.7ys.Many of you wasted time and energy computing the x squared and y squared columns and then decided that you could get the xy sum by multiplying the sums of x and y. Where have you been?b. The sample covariance between x and y. (3)c. The sample correlation between x and y. (2)d. Given the size and sign of the correlation, what conclusion might you draw on the relation between speed and safety if this were the only evidence available? (1)e. Assume that the death rate in all 9 countries fell by .1. What would be the new values of ,x ,xsxys and xyr. Use only the values you computed in a-c and rules for functions of x and y to get your results. If you state the results without explaining why, or change x and recompute the results, you will receive no credit. (4). How many of you recomputed the results anyway?Solution: a) 24444.492.38nxx 9653.0872256.7824444.4986.16912222nxnxsx 9825.09653.0 xs.00.609540nyy  25.568450800.6093285012222nynysy50.725.56 ysb) We found above that 2326xy , so  25.4834800.6024444.49232612nyxnxysxy c)   5768.50.79825.025.4yxxyxysssr This must be between -1 and 1. d) If we square the correlation we get 0. 333, which in a zero to one scale is not impressive. I would not want to conclude that speed limits and safety are closely related, though speed limits may be a factor.3251y0341 05/05/03e) From the syllabus supplement article: “Let us introduce two new variables, w and v, so that baxw , and dcyv , where ,,, cba and d are constants. From the earlier part of this section we know the following:      bxaEbaxEwEw    2222xwaxVarawVar      dycEdcyEvEv    2222yvcyVarcvVar To this we now add a new rule:    xywvacyxacCovvwCov ,, To find the correlation between w and v , recall that vwwvwv . But since 222xwa and 222yvc , then   xyyxxyyxxyyxxywvacsignacacacaccaac2222 .” Since these rules work for sample statistics too, then 1.0xw, yv  , so ,1a,1.0b 1c and .0d 24444.41.24444.41. xu 22221xxwsss  , so50.725.56 ws.   25.411 xywvss.  5768.2222xyyxxyyxxyyxxywvracsignsssacacssacacsscsaacsr 4251y0341 05/05/03III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points - Anything extra you do helps, and grades wrap around) . 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! If you are using the


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