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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 11.5 Homework Answers
Homework: pg. 506, #'s 2, 8. 2. We have that X = (X1 + + X35 )/35. Let Z be a standard normal RV. Using the central limit theorem gives P (460 < X < 540) = P X1 + + X35 < 540 35 X1 + + X35  35 500 = P 40 < < 40 35 X1 + + X35  35 500 < 2.366 = P 2.366 < 35 100 P (2.366 < Z < 2.366) 460 < 1 = 2 = 0.982.
2.366 ex
2.366 2 /2 dx 8. Let Xi be the number of people that the ith employee brings to the party (i {1, 2, . . . , 300}). Then the probability mass function of each Xi is given by p(0) = P {Xi = 0} = 1/3 p(1) = P {Xi = 1} = 1/3 p(2) = P {Xi = 2} = 1/3. Thus, E[X] = 0(1/3) + 1(1/3) + 2(1/3) = 1 and E[X 2 ] = 1(1/3) + 4(1/3) = 5/3. Therefore, V ar(X) = 5/3  1 = 2/3 and = 2/3. Let X = X1 + X2 + + X300 and let Z be a standard normal RV. Then using the central limit theorem gives P (X 320) = P (X1 + X2 + + X300 320) = P (X1 + X2 + + X300  300 1 20) X1 + X2 + + X300  300 2) = P( 300 2/3 P (Z 2) 1 2 = ex /2 dx 2 2 = 0.07865. 1 ...
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This note was uploaded on 06/24/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.
 Spring '08
 Anderson
 Central Limit Theorem, Probability

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