HWKsolutions4 - STAT35AHW4Solutions...

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1 STAT 35A HW4 Solutions http://www.stat.ucla.edu/~dinov/courses_students.dir/08/Spring/STAT35.dir Problem 1 a. E(X), the expected value of the R.V. X. E(X) = 0.08 × 0 + 0.15 × 1 + 0.45 × 2 + 0.27 × 3 + 0.053 × 4 = 2.06 b. V(X), the variance of X. V(X) = 0.08 × (0 2.06) 2 + 0.15 × (1 2.06) 2 + 0.45 × (2 2.06) 2 + 0.27 × (3 2.06) 2 + 0.053 × (4 2.06) 2 = 0.9364 c. SD(X), the standard deviation of X. SD(X) = 0.9364 ½ = 0.9677 d. V(X), using the shortcut formula, V(X) = E(X 2 ) μ 2 . Should equal the answer in part (b). V(X) = 0.08 × 0 2 + 0.15 × 1 2 + 0.45 × 2 2 + 0.27 × 3 2 + 0.053 × 4 2 – 2.06 2 = 0.9364 Problem 2 a. Compute the first three moments of the random variable X, E(X), E(X 2 ), E(X 3 ), and V(X). E(X) = 0.2 × 13.5 + 0.5 × 15.9 + 0.3 × 19.1 = 16.38 E(X 2 ) = 0.2 × 13.5 2 + 0.5 × 15.9 2 + 0.3 × 19.1 2 = 272.298 E(X 3 ) = 0.2 × 13.5 3 + 0.5 × 15.9 3 + 0.3 × 19.1 3 = 4592.276 V(X) = 272.298 – 16.38 2 = 3.9936 b. If the price of a freezer having capacity X ft3 is 25X 8.5, what is the expected price paid by the next customer to buy a freezer? E(25X 8.5) = 25 × 16.38 – 8.5 = 401 c. What is the variance of the price 25X 8.5 paid by the next customer? Interpret this value!
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HWKsolutions4 - STAT35AHW4Solutions...

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