# HW1a(1) - Econ 4010 Introduction to Econometrics(Spring...

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Econ 4010: Introduction to Econometrics (Spring 2016) Homework 1 Answers Olivier Parent ( [email protected] ) Assignment 1. Find the following probabilities assuming a normal distribution, with a mean of 7 , 450 and a standard deviation of 300: (a) P ( x 7 , 000); P ( x 7 , 000) is equal to: P ( x - μ σ 7 , 000 - μ σ ) P ( z ≤ - 1 . 50) = P ( z > 1 . 50) = . 5 - . 4332 = . 0668 (b) P ( x 8 , 000); P ( x 8 , 000) is equal to: P ( x - μ σ 8 , 000 - μ σ ) P ( z > 1 . 83) = . 50 - . 4664 = . 0336 2. Find the following probabilities assuming a normal distribution, with a mean of 7 , 450 and a standard deviation of 300: (a) P ( x 7 , 000); P ( x 7 , 000) is equal to:

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P ( x - μ σ 7 , 000 - μ σ ) P ( z ≤ - 1 . 50) = P ( z > 1 . 50) = . 5 - . 4332 = . 0668 (b) P ( x 8 , 000); P ( x 8 , 000) is equal to: P ( x - μ σ 8 , 000 - μ σ ) P ( z > 1 . 83) = . 50 - . 4664 = . 0336 (c) P ( x 8 , 250), P ( x 8 , 250) is equal to: P ( x - μ σ 8 , 250 - μ σ ) P ( z > 2 . 67) = 50 - . 4962 = . 0038 (d) P (7 , 400 x 7 , 700) P (7 , 400 x 7 , 700) is equal to: P ( 7 , 400 - μ σ x - μ σ 7 , 700 - μ σ ) P (7 , 400 < x < 7 , 700) = P ( - 0 . 17 < x < 0 . 83) = . 0675+ . 2967 = . 3642 3. A distribution has a normal distribution with a mean of 109 and a standard deviation of 23,5. (a) Calculate the probability of a value being less than 101.3; P ( x 101 . 3) is equal to: P ( x - μ σ 101 . 3 - μ σ ) P ( x < 101 . 3) = P ( z < - 0 . 33) = P ( z > 0 . 33) = . 50 - . 1293 = . 3707
(b) Calculate the probability of a value being more than two standard deviations above the mean; two standard deviations above the mean = 109 + 2(23 . 5) = 156 P ( x 156) is equal to: P ( x - μ σ 156 - μ σ ) = 0 . 50 - 0 . 4772 = 0 . 0228 4. Let X 1 stand for the rate of return on one security, and let X 2 stand

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• Spring '16
• Jeff Mills

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