(c) The continuous random variable Y with probability density function f(y) = 9y for 0 le y le 1, density zero elsewhere - M_Y(t) = 9left(frac{e^t(t - 1) + 1}{t^2} ight). - M_Y(t) = 10left(frac{e^t(t - 1) + 1}{t^2} ight). - M_Y(t) = 10(e^t(t - 1) + 1). - M_Y(t) = 9(e^t(t - 1) + 1).
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This is given by: E(Y) = ∫y*f(y) dy, where f(y) is the probability density function of Y. Substituting the given density function, we get: E(Y) = ∫0^1 y*9y dy = 9∫0^1 y^2 dy = 9*[y^3/3]0^1 = 3 Therefore, the expected value of Y is 3. Show more…
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