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This best-selling engineering information textual content presents a pragmatic technique that's extra orientated to engineering and the chemical and actual sciences than many related texts. it truly is choked with specific challenge units that mirror sensible occasions engineers will come upon of their operating lives.
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5112. 3XO P (green ball) = 0. 128. 12 (a) (b) (c) (d) P(bag 1) x P(b1ue ball I b&g 1) P(b1ue ball) P(10) = P(score on die is 5) x P(tai1s) = Q x 1 2 -1 12P(3) = P(score on die is 3) x P(heads) = Q x f = P(6) = P(score on die is 6) (P(score on die is 3) x P(tails)) = Q + (QXf)=$. A. 16 : S = (-1,2,'3,4,5,6, 10) ( 4 0. 5112 = P(score on die is odd n 6 is recorded) P(6 is recorded) (# 5* = 625 45 = 1,024 In this case 54 < 45 and in general, n? < = ny 3. when 3 5 nl < nz. Let A be the event that the order is from a first time customer and let B be the event that the order is dispatched within one day.
X n2) = ( 6 Therefore Var(X) = E ( x 2 ) - (E(x))~ = - ( Y )=~12 6 (b) E ( X ) = (c) ' . (a) In this case X = XI +Xz, the sum of the times of the two rides. The expectation is E ( X ) = E (XI) E(X2) = 87 87 = 174 minutes. 24 minutes. 14 + . ,,,... - + + + (b) In this case X = 2 x XI, twice the time of the ride. The expectation is E ( X ) = 2 x E(X1) = 2 x 87 = 174 minutes. The variance is Var(X) = 22 x Var(Xl) = = 6 minutes. 22 x 32 = 36, and the standard deviation is . l. I\<,,. :.. ,, . jli. :: I.
6. 3 (a) xi 1 2 3 4 5 CHAPTER2. RANDOM' U Let Xi be the value of the ith card dealt. )' for 1 1x -- + ... 6' 5 11. 332. 706. 305. 599. 6 (a) fx(x) = (b) fi(y) = 4x(2-y) dy = 2 1 for 0 5 x 5 1. 2. Since f (x, y) = fx(x) x fr(y) the random variables are independent. . (c) Cov(X, Y) = 0 since the random variables are independent. 7. 10 ' ! Also which is 1 if a > 0 and -1 if a < 0. L CHAPTER 2. 11 The expected amount of a claim is 1 11800 E(X) = ~(1,800 2) 972,000,000 0 dsc = $900. 1 x $900) = $5.
