# 2 - STAT 350 Fall 2008 Homework #2 Solution through Lecture...

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STAT 350 – Fall 2008 Homework #2 – Solution through Lecture #4 1. Let X ~ Normal( μ = 12, σ = 3) and Y ~ Normal( = -2, = 4). X and Y are independent random variables. Let W = 8 – 6 X + 2 Y . a. Find the distribution of W . You can think of W as the sum of (8 – 6X) and (2 Y ). (8 – 6 X ) is Normal with mean = 8 – 6(12) = -64 standard deviation = |-6|(3) = 18 (2 Y ) is Normal with mean = 2(-2) = -4 standard deviation = |2|(4) = 8 Thus [(8-6 X ) + (2 Y )] is Normal with mean = -64 + (-4) = -68 standard deviation = 22 18 8 388 19.698 += W ~ Normal( = -68, = 19.698) b. Find P( W < -50) () ( 68 50 68 0.91 0.91 19.698 19.698 W PP Z −− − −− ⎛⎞ <= < ⎜⎟ ⎝⎠ ) = Φ = 0.8186 2. Assume that the weight in pounds of pigs on a particular farm is normally distributed with mean of 200 pounds and standard deviation of 25 pounds. For parts (a) through (e): Let X = the weight in pounds of a (single) randomly selected pig from this farm X ~ Normal( = 200, = 25) a. What is the probability that a randomly selected pig from this farm weighs less than 190 pounds? ( 200 190 200 190 0.40 0.40 25 25 X PX P PZ < = < = Φ ) = 0.3446 b. What is the probability that a randomly selected pig from this farm weighs more than 220 pounds? ( 200 220 200 220 0.80 1 0.80 25 25 X P >= > = > = Φ ) = 1 – 0.7881 = 0.2119 Homework #2 – Solution Page 1 of 5

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c. What is the probability that a randomly selected pig from this farm weighs between 190 and 220 pounds?
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## This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue.

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2 - STAT 350 Fall 2008 Homework #2 Solution through Lecture...

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