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Statistics Homework Solutions 91

Statistics Homework Solutions 91 - 91 SECTION 4.6 0 02 and...

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SECTION 4.6 91 5. (a) ln I N 1 0 2 , ln R N 4 0 1 , and I and R are independent. Therefore ln V N 5 0 3 . Since ln V is normal, V is lognormal, with μ V 5 and σ 2 V 0 3. (b) P V 200 P ln V ln200 P ln V 5 298317 . Now ln V N 5 0 3 , so the z -score of 5.298317 is 5 298317 5 0 3 0 54. The area to the left of z 0 54 is 0.7054. Therefore P V 200 0 7054. (c) P 150 V 300 P ln150 ln V ln300 P 5 010635 ln V 5 703782 . Now ln V N 5 0 3 , so the z -score of 5.010635 is 5 010635 5 0 3 0 02, and the z -score of 5.703782 is 5 703782 5 0 3 1 28. The area between z 0 02 and z 1 28 is 0 8997 0 5080 0 3917. Therefore P 150 V 300 0 3917. (d) The mean of V is E V e μ σ 2 2 e 5 0 3 2 172 43. (e) Since ln V N 5 0 3 , the median of ln V is 5. Therefore the median of V is e 5 148 41. (f) The standard deviation of V is e 2 μ 2 σ 2 e 2 μ σ 2 e 2 5 2 0 3 e 2 5 0 3 101 99. (g) Let v 10 be the 10th percentile of V . Then P V v 10 P ln V ln v 10 0 10. The z -score of the 10th percentile is approximately z 1 28. Therefore the z -score of ln v 10 must be
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