Discrete-Solu - Answer Key for Discrete Problems 1. It is...

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Answer Key for Discrete Problems 1. It is estimated that one out of 10 widgets supplied to a manufacturer is defective. Suppose a random sample of 20 is chosen from a very large batch of these widgets. What is the probability that the sample contains no more than one defective widget? Assume that the number of defectives follows a binomial distribution. We want P(X = 0 or 1) = P(X = 0) + P(X = 1) = 1*0.1 0 *0.9 20 + 20*0.1 1 *0.9 19 = 0.1216 + 0.2701 = 0.3917 2. A manufacturer of heating elements for water heaters ships boxes that contain 100 elements. A quality-control inspector randomly selects a box in each shipment and accepts the shipment if there are 5 or fewer defective heating elements in the box. Assuming that the manufacturer has had a rate of 6% defective items, what is the probability that the shipment of heating elements will pass the inspection? Let D = # defectives in a box. We can model D with a binomial distribution with n = 100, p = 0.06. We want P(D <= 5) Use BINOMDIST(5,100,0.06,1) = 0.4407. Alternatively using P(X = k) = nCkp k (1-p) n-k , P(X = 5) = 100C5*.06 5 *.94 95 = (100*99*98*97*96/5*4*3*2*1)* 0.00000078* 0.0027999 = . 16392 P(X = 4) = 100C4*.06 4 *.94 96 = (100*99*98*97/(4*3*2*1)* 0.00001296*0.002631928 = .13375 P(X = 3) = 100C3*.06 3 *.94 97 = (100*99*98/3*2*1)*0.000216*0.002474012 = .08641 P(X = 2) = 100C2*.06 2 *.94 98 = (100*99/2*1)*0.0036* 0.00232557 = .04144 P(X = 1) = 100C1*.06 1 *.94 99 = (100/1)*0.06*0.002186 = .01312 P(X = 0) = 100C0*.06 0 *.94 100 =1*0.002* 1 = .00205
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P(X <=5) = sum of above = .16392 + .13375 + .08641 + .04144 + .01312 + .00205 = 0.44069 ~
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Discrete-Solu - Answer Key for Discrete Problems 1. It is...

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