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WCU ECO 251 - ECO 251 Third Exam

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251y0831 4/18/08 ECO251 QBA1 THIRD EXAMApr 24, 2008 Name: ________KEY_________ Student Number: _____________________Class time: _____________________ Part I. (16 points) Do all the following (2 points each unless noted otherwise). Make Diagrams! Show your work! In particular you must briefly explain how you got the answer to the value of z at the bottom ofthis page.zhas the standardized Normal distribution  1,0~ Nz for the first four problems.1.  23.1zP   8907.5.3907.0023.1  zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the entire area above -1.23. Because this is on both sides of zero, we must add the area between -0.75 and zero to the area above zero.2.  025.3  zP4994.For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade only the area between -3.25 and zero. Because this is on one side of zero and ends at zero, we can simply look up 3.25 on the table. The area between -3.25 and zero is exactly the same as the area between zero and 3.25 if the mean is zero. The bottom of our Normal table has the following.For values above 3.09, see belowIf z0 is between P z z00  is 3.08 and 3.10 .4990 3.11 and 3.13 .4991 3.14 and 3.17 .4992 3.18 and 3.21 .4993 3.22 and 3.26 .4994 3.27 and 3.32 .4995 3.33 and 3.38 .4996 3.39 and 3.48 .4997 3.49 and 3.61 .4998 3.62 and 3.89 .4999 3.90 and up .5000Since 3.21 is between 3.22 and 3.25 we use .4994.3.  07.307.3  zP   9978.4989.4989.07.30007.3  zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -3.07 and 3.07. Because this is on both sides of zero, we must add the area between-3.07 and zero to the area between zero and 3.07. These areas are identical.4. 135.z Solution: 135.z is, by definition, the value of z with a probability of 13.5% above it. Make a diagram. The diagram for z will show an area with a probability of   %5.1310086.5% below135.z. It is split by a vertical line at zero into two areas. The lower one has a probability of 50% and the upper one a probability of 86.5% - 50% = 36.5%. The upper tail of the distribution above 135.z has a probability of 13.5%, so that the entire area above 0 adds to 36.5% + 13.5% = 50%. From the diagram, we want one point 135.z so that  5.86.135.zzP or  3650.0135. zzP. If we try to find this point on the Normal table, the closest we can come is  3646.10.10 zP, though 3665.11.10 zP is more or less acceptable. . So we will use 10.1135.z, though 1.11 might be acceptable.1251y0831 4/18/08 7,4~ Nx for problems 5 through 8. Note that all values of z are rounded to the nearest hundredth.On this page we must make the simple transformation xzto ‘standardize’ our values of x.5.  23.1xP 75.07423.1 zPzP   7734.5.2734.0075.0  zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the entire area above -0.75. Because this is on both sides of zero, we must add the area between -0.75 and zero to the area above zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area above -1.23. This area is on both sides of the mean (4), so we add the area between -1.23 and the mean to the whole area (.5) above the mean .6. 025.3  xP 57.004.17407425.3 zPzP   1351.2157.3508.057.0004.1  zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -1.04 and -0.57. Because this is entirely on one side of zero, we must subtract the smaller area between -0.57 and zero from the larger area between -1.04 and zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area between -3.25 and 0. Both these numbers are to the left of the mean. This areais on one side of the mean (4), so we subtract the smaller area between -3.25 and the mean from the larger area between 0 and the mean.7.  07.307.3  xP 13.001.17407.37407.3 zPzP   2921.0517.3438.013.0001.1  zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the zero by a vertical line! Shade the area between -1.01 and -0.13. Because this is entirely on one side of zero, we must subtract the smaller area between -0.13 and zero from the larger area between -1.01 and zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 4. Indicate the mean by a vertical line! Shade the area between -3.07 and 3.07. Both these numbers are to the left of the mean. This area is on one side of the mean (4), so we subtract the smaller area between 3.07 and the mean from the larger area between -3.07 and the mean.8. 135.x Solution: 135.x is, by definition, the value of x with a probability of 13.5% above it. If you did not do so on the last page, make a diagram. The diagram for z will show an area with a probability of  %5.1310086.5% below 135.z. It is split by a vertical line at zero into two areas. The lower onehas a probability of 50% and the upper one a probability of 86.5% - 50% = 36.5%. The upper tail of the distribution above 135.z has a probability of 13.5%, so that the entire area above 0 adds to 36.5% + 13.5% = 50%. From the diagram, we want one point 135.z so that  5.86.135.zzP or 3650.0135. zzP. If we try to find this point on the Normal table, the closest we can come is 3646.10.10 zP, though  3665.11.10 zP is more or less acceptable. . So we willuse 10.1135.z, though 1.11 might be acceptable.2251y0831 4/18/08You have gotten to this point


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