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Continuous Distributions

# Continuous Distributions - 28 Continuous Functions Moments...

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28 Continuous Functions Moments 1. The k th moment about the origin of the distribution of the continuous random variable X whose pdf is f is given by: = b a k k dx x f x ) ( ' μ for domain, a x b For k = 1, μ is the theoretical mean. 2. The k th moment about the mean of the distribution of the continuous random variable X whose pdf is f is given by: - = b a k k dx x f x ) ( ) ( μ μ for domain, a x b For k = 1: 0 1 = μ , always. For k = 2: , 2 2 δ μ = the variance. And 2 ' 2 2 μ μ δ - = . For k = 3: 3 μ measures skewness. For k = 4: 4 μ measures kurtosis. Expectation = = b a dx x xf X E ) ( ) ( μ , where f(x) is a pdf. OR = b a dx x f x g X g E ) ( ) ( )] ( [ . Example : Given: 2 ) ( cx x f = , 0 < x < 2. Find: a) c to make f a pdf. b) the mean of X . c) the variance of X . a) 8 3 1 3 8 3 1 1 2 0 3 2 0 2 0 2 2 = = = = = c c x c dx x dx cx b) 2 3 32 3 8 3 8 3 2 0 2 0 4 3 2 2 0 = = = x dx x dx x x c) = + - = - 2 0 2 3 4 2 2 2 0 20 3 ) ) 4 / 9 ( 3 ( 8 3 8 3 2 3 dx x x x dx x x

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29 Chebyshev's Theorem Let k be any number such that k > 1. The proportion of observations in any population or sample lying within a distance of k standard deviations from the mean is at least 2 1 1 k - . Example : If k = 2, then 4 3 2 1 1 2 = - of the data points, or more, will lie in the interval δ μ 2 ± . Only 1/4 or less of the data points are in the right and/or left tails of the distribution. Example : On an I.Q. test with mean of 100 and a standard deviation of 15, there will be at least 3/4 of the scores between 70 and 130 and that there will be at least 8/9 of the scores between 55 and 145. Rectangular Distribution : b x a a b x f - = , 1 ) ( Example : Two students agree to meet at a restaurant between 6 and 7 p.m. Find the probability that they will meet if each agrees to wait 15 minutes for the other and they arrive independently at random between 6 and 7. Sketch a square and place student A on the horizontal and student B on the vertical axis. Label the bottom left corner 6:00 pm for both and the top right corner represents 7:00 pm for both. If the students arrive at the same time, that time would be on a diagonal from the bottom left corner to the top right corner. On either side of this diagonal, for 15 minutes (1/4 of an hour), the two students would meet. The remaining triangles, one in the lower right and the other in the upper left, equal in size, represent the area when the two students would not meet. The area of the square is one hour by one hour. _____ _____ _____ _____ 7:00 | | Given that the area of the square is | | one, find the area of the two triangles student B | | to get the area when the students will | | not meet. One triangle is: | | hb A 2 1 = = 32 9 4 3 4 3 2 1 = |_____________________| and 2(9/32) = 9/16, the area of both, 6:00 7:00 the probability that they will NOT student A meet. So, 1 - 9/16 = 7/16 is the pro- bability that the students will meet.
30 Normal Distribution (Gaussian Distribution) The function, 2 2 1 2 1 ) ( - - = δ μ π δ x e x f generates the family of normal distributions, with parameters of μ and δ .

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