Quiz2-2007-2- test 2- 610 question 1

5767 5 sol let x denote the amount of cappuccino the

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Unformatted text preview: 1/10). That is, P (X1 + X2 + X3 > 90) = P (N90 ≤ 2) 2 = 9x e−9 x! x=0 = e9 + 9e−9 + = 0.006263 92 e−9 2 11. (6%) The cappuccino machine at the Cambridge Inn can be adjusted to dispense about µ ounces (oz) each time, for any 0 < µ < 10, but the actual amount dispensed is variable with a standard deviation of 0.1 oz. If the amounts dispensed follow a normal probability distribution, how should µ be set to ensure that only one 6 oz cup in a hundred overflows, on average? µ = 5.767 5 Sol) Let X denote the amount of cappuccino the machine output. Then X ∼ N (µ, 0.12 ). We want to determine the value of µ such that P (X > 6) = 0.01. 6−µ 0.1 0.01 = Φ(−2.33) = P (X > 6) = 1 − Φ =⇒ 10µ − 60 = = 1 − Φ(60 − 10µ) = Φ(60µ − 10) −2.33 µ = 5.767 12. (4%) Let (X, Y, Z ) have joint probability density function 4x for 0 < x < y < z < 1 f (x, y, z ) = 0 otherwise. Find P ((X + Z )/2 ≥ Y ). (Hint: Do the integration in the order 4xdydxdz ). Sol) 1 P ((X + Z )/2 ≥ Y ) z (x+z )/2 = 4xdydxdz 0 0 1 x z = 0 0 1 2x(z − x)dxdz 0 0 1 2 z x2 − x3 3 = 0 1 = 0 = 6 dydxdz x z = = (x+z )/2 4x z4 12 1 12 z3 dz...
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