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111 . AA, E --|9- ge wrap Text Cut la Copy Paste-t For mat Painter Calibri (Body) a-A- ???!???Merge & Center ?? Font Alignment Clipboard TextBox 1 ?| 4 decim -Cum Similar to the practice activity, we are going to figure out our expected and requested budgets for repairing our 16 widgets next fiscal year. At present, we are taking 5sample sizes of 60 widgets each, and we have a known 6 | defective rate of 16.2%; this means that we expect an 7 average of 9.7 widgets per sample to be rejects. 9Consequently, if 10 or fewer widgets are rejected, for whatever reason, then we will say that our quality is good enough and we will pass that production lot. Orn 11the other hand, if we find 11, 12, or 13 bad widgets in 12 our 60, then, well, may-be our quality is good enough or 13 may-be it is not, so we need to double check and we will 14 pull a second sample of 60 widgets. Then again, 14 or 10 more defective widgets in the first sample means stop what we are doing and fix things. 12 17 IF we pull a second sample, then we will again pass the o 18 lot if we find 10 or fewer defective units; else, we will 19 reject that lot. Based on this information, fill in the cells at the right as 17 21 22 indicated and directed (see PA) 2You will need this information 26 #Expected Production Lots 2,013 28 Average Production Lot Size 894 widgets Expected Cost to Repair a Widget $2.53 32 33 0 31 37 Filtering Data Binomial Normal Distr AppExplanation / Answer
Solution
Let X be the number of defectives in a random sample of size n taken from a lot with proportion of defectives = p. Then, X ~ B(n, p).
Back-up Theory
If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n ………….………..(1)
[The above probability can also be directly obtained using Excel Function of Binomial Distribution: BINOMDIST(Number_s:Trials:Probability_s:Cumulative), what is within brackets is (x:n:p:True)] ……………………………………………………..………….(1a)
Now to work out the solution,
Given first sample n1 = 60, p = 0.162 (16.2%) and lot is accepted for 10 or less defectives, second sample of n2 = 60 is taken for 11, 12 or 13 defectives and lot is rejected for 14 or more defectives,
Pr(pass lot/I sample) = P(X ? 10)
= 0.621583 [vide (1a) under Back-up Theory] ANSWER 1
Pr(II sample /I sample) = P(11 ? X ? 13)
= 0.28192 [vide (1a) under Back-up Theory] ANSWER 2
Pr(fail lot/I sample) = P(X ? 14)
= 0.096497 [vide (1a) under Back-up Theory] ANSWER 3
Further given for second sample, lot is accepted for 10 or less defectives, and lot is rejected for 11 or more defectives,
Pr(pass lot/II sample) = P(X ? 10)
= 0.621583 [vide (1a) under Back-up Theory] ANSWER 4
Pr(fail lot/II sample) = P(X ? 11)
= 0.378417 [vide (1a) under Back-up Theory] ANSWER 5
Pr(overall passing the lot) = 0.621583 + (0.28192 x 0. 621583) = 0.79682 ANSWER 6
Pr(overall failing the lot) = 0.096497 + (0.28192 x 0.378417) = 0.20318 ANSWER 7
In the long run, proportion of lots that will pass = 0.79682 or 79.682% ANSWER 8
In the long run, proportion of lots that will fail = 0.20318 or 20.318% ANSWER 8
DONE
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