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Calculus and Beyond Homework Help
Find the standard deviation of the values of ##y##
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[QUOTE="Mark44, post: 6855598, member: 147785"] Sorry, I made a typo (now corrected) in one of the formulas I wrote: It should be Var(X + a) = Var([B]X[/B]), not Var(a) as I originally wrote. So if U = 3x, Var(U) = Var(3x), and so Var(U + 1) = Var(U). No, that's wrong on several counts. ##Var (Y) \ne \sqrt {a^2×Var(3x)+1}## -- you're mixing up variance and standard deviation. Var(Y) = Var(3X) = 3[SUP]2[/SUP] Var(X) So SD(Y) = ##\sqrt{9 Var(X)} = 3 \sqrt{Var(X)}## The answer given in the image you posted, 2.64, is correct. 2.82 is not. It doesn't affect the calculation of the variance. A vertical translation in the data makes no difference. For example, the sets {1, 2, 3} and {3, 4, 5} have different means but the same variance, and hence the same standard deviations. [/QUOTE]
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Calculus and Beyond Homework Help
Find the standard deviation of the values of ##y##
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