midterm2+fall+2009+solutions

midterm2+fall+2009+solutions - ECI 114 Fall 2009 Name...

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Unformatted text preview: ECI 114 Fall 2009 Name Midterm 2 (100 points) Wednesday, November 18, 2009 Show your work. It’s not necessary to give a final numerical answer unless specifically asked, but you need to provide all the information necessary to compute the final answer. Please circle or otherwise highlight your answer. Turn in your “crib sheet” and your normal table with your exam. Put your name on the back of the last page. GOOD LUCK! 1. For each of the following statements, circle the letter “T” if it is true, and “F” if it is false. (3 * 8 =24 points) T F a. Let Φ (x) denote the CDF of the standard normal distribution; then Φ (- 0.2) = - Φ (0.2). False. Φ (- 0.2) = 1 - Φ (0.2). In addition, the right side of the equation, “- Φ (0.2)”, becomes negative. Since the probability must not be less than 0, the equation is incorrect. T F b. If a random variable Y is distributed with pdf f(y) = 0.25 for 0 ≤ Y < 4, then the probability that Y = 1 is 0.25. False. The probability that Y = 1 is zero. For a continuous random variable, Pr(Y=1) ≠ f(1) = 0.25. See page Ran-12 of the notes, “f(y) is a density, not a probability”. T F c. For a continuous random variable Y and any given real numbers a and b, Pr[a < Y < b] = Pr[a ≤ Y ≤ b]. True. For a continuous random variable Y, the probability that Y equals a single real value is always zero. Therefore, Pr[a ≤ Y ≤ b] = Pr[a < Y < b] + Pr[Y = a] + Pr[Y = b] = Pr[a < Y < b] T F d. If X 1 , X 2 , …, X n identically and independently follow the Bernoulli distribution with parameter p, then ∑ = n i i X 1 ~ bin(n,p). True. For a R.V. X with a Bernoulli distribution, X = 1 represents that the Bernoulli trial “succeeds”, while X = 0 represents that the Bernoulli trial “fails”. If X 1 , X 2 , …, X n identically and independently follow the Bernoulli distribution with parameter p, then ∑ = n i i X 1 equals the number of “successes” out of n Bernoulli trials. Therefore, ∑ = n i i X 1 ~ bin(n,p) T F e. The geometric distribution is a special case of the binomial distribution. False. The geometric distribution is a special case of the negative binomial distribution Midterm 2, ECI 114, FALL 2009 Page 1 T F f. If Y = 2X + 1, where X~N(0,1), then Y~N(1,4). True. E(Y) = 2*E(X) + 1 = 2*0 + 1 = 1. Var(Y) = 2 2 *Var(X) = 4*1 = 4. A linear transformation of any normally distributed R.V. still follows a normal distribution. Therefore, Y~N(1,4). T F g. For both discrete random variables and continuous random variables, the CDF F(a) ≤ F(b) ⇒ a ≤ b. False. For example, if Pr( 1 < X ≤ 2) = 0, then F(2) = Pr(1 < X ≤ 2) + F(1) = F(1). Thus, F(2) ≤ F(1) but 2 > 1. It would have been true if we’d said “F(a) < F(b) ⇒ a < b”. F(a) = Pr(X ≤ a), and F(b) = Pr(X ≤ b). F(a) < F(b) ⇒ Pr(X ≤ a) < Pr(X ≤ b). If a = b, then F(a) = F(b) , which contradicts the given information that F(a) < F(b). If a > b, then Pr(X ≤ a) = Pr(X ≤ b) + Pr(b < X ≤ a). Because Pr(b < X ≤ a) ≥ 0, Pr(X ≤ a) ≥ Pr(X ≤ b), which ≤ b) + Pr(b < X ≤ a)....
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midterm2+fall+2009+solutions - ECI 114 Fall 2009 Name...

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