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Course: MATH 331, Spring 2008
School: Wisconsin
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Spring Math331, 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 &lt; X &lt; 540) = P X1 + + X35 &lt; 540 35 X1 + + X35 - 35 500 = P -40 &lt; &lt; 40 35 X1 + + X35 - 35 500 &lt; 2.366 = P -2.366 &lt; 35 100 P...

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Spring Math331, 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 e-x -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 = e-x /2 dx 2 2 = 0.07865. 1
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