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Unformatted text preview: ECO 230 Marianna Kudlyak Economic Statistics Spring 2007 Answer Key to Homework #7 Total points: 10. Graded exercises: Total points: 10. Graded exercises: Exercise 2 (1.5 points),3 (1.5 points), 7(2 points), and 8(3 points). Half point was added for the presence of each of the remaining 4 exercises. Exercise 1 Consider the following probability distribution function: f(x) = 1/c for 3 < x < 7 = 0 otherwise a) Find c so that f(x) is a proper probability distribution function and draw the probability distribution function. f(x) is uniform pdf on the interval (a,b). For f(x) to be proper pdf, the following properties should hold: ∫ ∫ ∈ ≥ = ∞ ∞ − v u R v u for dx x f dx x f , ) ( . 2 1 ) ( . 1 Hence, to find c we integrate 1/c on the interval from a to b and equate it to 1; c = 7( 3)=10 Since 10> 0 , graph of f(x) lies above xaxis. Thus, we conclude that second property holds. Answer: c = 10 b) Find and draw the cumulative distribution function. F(x ) = 0 if x 0 < 3 F(x ) = 0 if x 0 > 7 10 ) 3 ( 10 1 ) ( − − = = ∫ x dx x F x a if 3 <= x < = 7 3 7 1/10 x f(x) c) Find E(X). 2 2 7 3 10 * 2 ) 3 ( 7 10 1 ) ( 2 2 7 3 = + − = − − = = ∫ − dx x X E d) Find var(X). ( ) 12 100 10 * 3 ) 3 ( ) 7 ( 2 10 1 )) ( ( ) ( ) var( 3 3 2 7 3 2 2 2 = − − = − = − = ∫ − dx x X E X E X e) Find the median of X. P(X<=m) = 0.5 5 . 10 1 3 = ∫ − dx m After solving for m , we get m = (3+7)/2=2 f) Find standard deviation of Y, where Y = 2 – 5X 3 25 3 2 10 5 ) ( 5 ) var( 5 ) 5 2 var( ) var( ) ( = = = = − = = X X X Y Y σ σ . Exercise 2. Recall that Normal distribution is to satisfy the following requirements: the graph should be bell shaped curve; mean, median and mode are all equal; mean, median and mode are located at the center of the distribution; it has only one mode, it is symmetric about mean, it is a continuous function; it never touches xaxis; and the area under curve equals one. Let X ~N(50, 64). In the answers below I standardize the random variable by constructing new random variable ) 1 ; ( ~ N Z X Z σ µ − = a) Find the probability that X is greater than 60. P(X>60) = P(Z>1.25) = 1F(1.25) = 0.106 3 7 1 x F(x) b) Find the probability that X is greater than 35 and less than 62. P(35<X<62) = P(1.875<Z<1.5) = F(1.5) – F(1.875) = F(1.5) – (1  F(1.875)) = 0.903 c) Find the probability that X is less than 55....
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 Spring '08
 Kudlyak
 Normal Distribution, Standard Deviation, Newbold, probability distribution function, OO, ×Ó ÔÔÖÓÜ

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