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Unformatted text preview: Math 361 X1 Homework 9 Solutions Spring 2003 Graded problems: 1(a), 2, 4(b), 5(b) (3 points each  12 points maximal); 7 (Bonus problem): up to 2 additional points Problem 1. [4.R:25, variant] Suppose U is distributed uniformly on the interval (0 , 1), and let Y = min( U, 1 U ), and let Z = 2 Y . Find (a) the c.d.f. and the density of Y ; (b) the c.d.f. and the density of Z ; (c) E ( Y ) and V ar ( Y ). Solution. (a) Since min( U, 1 U ) is between 0 and 1 / 2 when U is between 0 and 1, the range for the density f Y ( y ) is the interval [0 , 1 / 2], and it suffices to compute f Y ( y ) for such values y . To get a formula for this density, we compute first the c.d.f. F Y ( y ) = P ( Y ≤ y ). We have F Y ( y ) = P ( Y ≤ y ) = P (min( U, 1 U ) ≤ y ) = P ( U ≤ y or 1 U ≤ y ) = P (0 ≤ U ≤ y or 1 y ≤ U ≤ 1) = ( y 0) + (1 (1 y )) = 2 y (0 ≤ y ≤ 1 / 2) . Thus, f Y ( y ) = F Y ( y ) = 2 (0 ≤ y ≤ 1 / 2) , which is the uniform distribution on (0 , 1 / 2). (b) Since Y has range [0 , 1 / 2], Z = 2 Y has range [0 , 1]. For 0 ≤ z ≤ 1, the c.d.f. of Z is F Z ( z ) = P (2 Y ≤ z ) = P ( Y ≤ z/ 2) = F ( z/ 2) = z , so the density of Z is f Z ( z ) = F ( z ) = 1 for ≤ z ≤ 1, i.e., Z is uniformly distributed on [0 , 1]. (c) By the formulas for the expectation and variance and the densities computed in (a), E ( Y ) = Z 1 / 2 y · 2 dy = 1 4 E ( Y 2 ) = Z 1 / 2 y 2 · 2 dy = 1 12 , V ar ( Y ) = E ( Y 2 ) E ( Y ) 2 = 1 12 1 4 2 = 1 48 . Problem 2. [4.1:7] Suppose the distribution of height over a large population of individuals is approximately normal. Ten percent of individuals in the population are over 6 feet tall, while the average height is 5 feet 10 inches. What, approximately, is the probability that in a group of 100 people picked at random from this population there will be two or more individuals over 6 feet 2 inches tall? Solution. This is a two stage problem. In the first stage, we compute the probability p = P ( X > 74), where X denotes the height of a randomly selected individual, measured in inches. We are given that X is (approximately) normally distributed, so (1) P ( X ≤ x ) ≈ Φ x μ σ , 1 where μ and σ are the mean and standard deviation of the normal distribution. We are also given that μ = 70 (since the average height is 5 feet 10 inches), and that P ( X > 72) = 0 . 1 (since 10% are over 6 feet tall). By (1), the latter condition implies 1 Φ 72 μ σ = 0 . 1 , or Φ(2 /σ ) = 0 . 9, using the value μ = 70. From the normal table, we get 2 /σ = 1 . 28 or σ = 2 / 1 . 28 = 1 . 56. Using (1) again with x = 74, μ = 72, and σ = 1 . 56, we obtain (2) p = P ( X > 74) ≈ 1 Φ 74 70 1 . 56 = 1 Φ(2 . 56) = 0 . 0052 ....
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This note was uploaded on 04/03/2008 for the course STAT 134 taught by Professor Aldous during the Spring '03 term at Berkeley.
 Spring '03
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