Solution: In both sections of the problem, , and .251x0531 4/18/05 (Preliminary!!!!) ECO251 QBA1 THIRD EXAMApr 25, 2005 Name: _____________________ Student Number: _____________________ Class Time (Circle) 1pm 2pmPart I: 16 points.z follows the standardized Normal distribution 1,0~ Nz. Find the following. Make diagrams! Add a vertical line at zero to these diagrams.1. 63.263.2 zP 9914.4957.263.20063.2 zPzPData AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between -2.63 and 2.63 is 0.99152. 63.211.3 zP 0034.4957.4991.063.2011.3 zPzPData AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area Between -3.11 and -2.63 is 0.00333. 11.311.3 zPF 9991.4991.5.11.300 zPzPData AxisDensity43210-1- 2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area to the Left of 3.11 is 0.99914. 115.z Solution: Make a diagram. 115.z is defined as a point with 11.5% above it and thus 100% - 11.5% =88.5% below it, so it is the 88.5 percentile. The diagram for z will show an area with a probability of 88.5% below115.z . It is split by a vertical line at zero into two areas. The lower one has a251y0531 4/22/05 (Preliminary!!!!) probability of 50% and the upper one a probability of 88.5% - 50% =38.5%. The upper tail of the distribution above115.z has a probability of 11.5%, so that the entire area above 0 adds to 50%. From the diagram, we want one point 115.z so that 8850.115.zzP or 3850.0115. zzP. The closest we can come is 3849.20.10 zP. We can say 20.1115.z. Data AxisDensity43210-1-2-3-4-50.40.30.20.10.0Normal Curve with Mean 0 and Standard Deviation 1The Area to the Right of 1.2 is 0.1151x follows the Normal distribution 2,63.2~ Nx. This means 63.2 and 00.2.Find the following. Make diagrams! Remember xz. As usual, people made diagrams of x with zero in the middle. Make up your mind! If you are diagramming x, put the mean in the middle; if you are diagramming z put zero in the middle.5. 63.263.2 xP 4957.063.2263.263.2263.263.2 zPzPData AxisDensity1050-50.200.150.100.050.00Data AxisDensity5.02.50.0-2.5-5.00.40.30.20.10.0Normal Curve with Mean 2.63 and Standard Deviation 2The Area Between -2.63 and 2.63 is 0.4957Normal Curve with Mean 0 and Standard Deviation 1The Area Between -2.63 and 0 is 0.49576. 63.211.3 xP 63.287.2263.263.2263.211.3 zPzP 0022.4957.4979.063.2087.2 zPzP2251y0531 4/22/05 (Preliminary!!!!) Data AxisDensity1050-50.200.150.100.050.00Data AxisDensity5.02.50.0-2.5-5.00.40.30.20.10.0Normal Curve with Mean 2.63 and Standard Deviation 2The Area Between -3.11 and -2.63 is 0.0022Normal Curve with Mean 0 and Standard Deviation 1The Area Between -2.87 and -2.63 is 0.00227. 11.311.3 xPF 24.00024.0263.211.3 zPzPzPzP5948.0948.5. Data AxisDensity1050-50.200.150.100.050.00Data AxisDensity5.02.50.0-2.5-5.00.40.30.20.10.0Normal Curve with Mean 2.63 and Standard Deviation 2The Area to the Left of 3.11 is 0.5948Normal Curve with Mean 0 and Standard Deviation 1The Area to the Left of 0.24 is 0.59488. 115.x In problem I4, we found 20.1115.z. The opposite of xz is zx , so115.x .03.5220.163.2115.zCheck: 20.1263.203.503.5 zPzPxP 1151.3849.5.20.100 zPzPData AxisDensity1050-50.200.150.100.050.00Data AxisDensity5.02.50.0-2.5-5.00.40.30.20.10.0Normal Curve with Mean 2.63 and Standard Deviation 2The Area to the Right of 5.03 is 0.1151Normal Curve with Mean 0 and Standard Deviation 1The Area to the Right of 1.2 is 0.11513251y0531 4/22/05 (Preliminary!!!!) Part II: (15+ points) Do all the following: All questions are 2 points each except as marked. Exam is normed on 50 points including take-home. Most questions come from Wonnacott and Wonnacott (1990) (Showing your work can give partial credit on some problems! In open-ended questions it is expected. Please indicate clearly what sections of the problem you are answering and what formulas you are using. Neatness counts!) Remember that you may not be able to finish this section, so ration your time on each problem. [Numbers in brackets are a cumulative total]1. If y tends to decrease as x increases, what can we say about the population correlation ? a)1.b)1.c)0.d) *0.2. The riskiness of a portfolio made up of two investments a) will be higher when the covariance is zero.b) will be higher when the covariance is negative.c) *will be higher when the covariance is positive.d) does not depend on the covariance. [2]Explanation: Remember from the portfolio material, if1 and 212211 PPRPRPR, then 2211REPREPRE and 2121222121,2 RRCovPPRVarPRVarPRVar is the variance of the return. Here 1P is the portion of $1 invested in stock 1, 2P is the portion of $1 invested in stock2, 1R is the return on stock 1 and 2R is the return on stock 2. The variance is a measure of riskiness, and the product of 1P and 2P must be positive. So the higher ,,21RRCov the higher RVar.3. Should read 70!!!!Fifteen items are randomly selected from a pilot production run of N items to check their quality. x is the number of defective items in the sample. The distribution of x can be considered approximately binomial if N is a) 15b) 152c) 400d) *1500e) All of the abovef) None of the above.Explanation: We can consider the population infinite if .20nN 70n and .1400702020 n The only number above 1400 is 1500.4. There are 24 Million people living in California. Approximately 8 million are in Los Angeles. The Census Bureau takes a random sample of 200 people from the state. Let x be the number of people in the sample who live in Los Angeles, then x is
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