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midterm2+fall+2010+solutions

midterm2+fall+2010+solutions - 1 ECI 114 Fall 2010 Name...

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ECI 114 Fall 2010 Name _____________________________ Midterm 2: Monday, November 15, 2010 Show your work and explain your answers as needed. It is not necessary to give a final numerical answer like “12” unless specifically asked; you just need to have the correct expression like “4!/2!”. Please circle or otherwise highlight your answer. Turn in your crib sheet and your normal table with your exam. Good luck! 1. For each of the following statements, circle the letter “T” if it is true, and “F” if it is false. (18 points) T F (a) The Bernoulli distribution is a special case of the geometric distribution. Bernoulli is a special case of the binomial distribution. Geometric governs the number of trials till the 1 st (next) success, and such a random variable can never take on the value 0, as the Bernoulli distribution does. T F (b) If X is distributed exp( λ ), then E(X)= λ . E(X) = 1/ λ for the exponential distributions. T F (c) If Y = 3X + 2, where X~N(0,1), then Y~N(2,9). E(Y) = 3E(X)+2 = 2 Var(Y) = 3 2 Var(X) = 9 Because linear combination of normal RVs are normally distributed, Y~N(2,9) T F (d) The negative binomial distribution assumes sampling without replacement. The negative binomial distribution involves Bernoulli trials, which are sampling with replacement. T F (e) If X ~ N(3,25), then Pr(X ≤ 2) = 0.42 (within roundoff). Pr(X≤2) = Pr( 5 3 2 5 3 - - X ) = Pr(Z≤-0.20) = 0.4207 T F (f) The exponential and chi-squared distributions are both special cases of the gamma family of distributions. T F (g) If X and Y are independent random variables, then Var(X-Y) = Var(X) - Var(Y). Var(X-Y) = 1 2 Var(X) +(-1) 2 Var(Y)= Var(X) + Var(Y). May not be very intuitive! But it is logical when you think about it long enough….
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