HW 5 Solution

HW 5 Solution - STAT 350 Assignment 5 Solutions(60 points...

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STAT 350 Assignment 5 Solutions (60 points) 26. (a) (2) x is a discrete random variable, so = xp x x () = (0)(.08) + (1)(.15) + (2)(.45) + (3)(.27) + (4)(.05) = 2.06. (b) (2) 2 = ( ) xp x x 2 = (0-2.06) 2 (.08) + (1-2.06) 2 (.15) + (2-2.06) 2 (.45) + (3-2.06) 2 (.27) + (4-2.06) 2 (.05) = .9364. The standard deviation of x is then = .9364 = .9677. 29. (a) (2) y is a discrete random variable, so py y must equal 1. Using the formula for p(y), we have k(1) + k(2) + k(3) + k(4) + k(5) = 1, or, 15k = 1, so k = 1/15. (b) (2) P(at most three forms required) = P(y 3) = p(1) + p(2) + p(3) = 1(1/15) +2(1/15) + 3(1/15) = 6/15 = 2/5 = .40, or 40%. (c) (2) = yp y y = 1(1/15) + 2(2/15) + 3(3/15) +4(4/15) + 5(5/15) = 55/15 = 11/3. (d) (2) 2 = ( ) yp y y 2 = (1-11/3) 2 (1/15) + (2-11/3) 2 (2/15) +(3-11/3) 2 (3/15) + (4-11/3) 2 (4/15) + (5-11/3) 2 (5/15) = 14/9 = 1.5555. Therefore, = 14 9 / = 1.2472. 31. Let x = the number of defective solder joints. n = 285 and =.01 (a) (2) Recall from Chapter 2 that: n for a binomial random variable. So,  85 . 2 01 . 285 Recall also that: 1 n So,     6797 . 1 99 . 01 . 285 (b) (2) For a PCB to be defect-free, all 285 solder joints must be defect-free. So: 05702 . 99 . 0 285 x P free defect are that PCBs all of proportion (or 0.0808 using the normal) (c) (2)

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HW 5 Solution - STAT 350 Assignment 5 Solutions(60 points...

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