Acceptance Sampling Random sampling from a large lot of manufactured items yields a number of defective items X with an approximate binomial

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Acceptance Sampling Random sampling from a large lot of manufactured items yields a number of defective items X with an approximate binomial distribution with p being the true proportion of defectives in the lot. A sampling plan consists of specifying the number, n, of items to be sampled and an acceptance number aa. After n items are inspected, the lot is accepted if X sa and is rejected if X>a. a. For n=5 and a=0 calculate the probability of accepting the lot for values of p equal to 0, 0.1, 0.3, 0.5 and 0.9 and 1 Write the six acceptance probabilities, separated by commas. Then write a brief explanation to support the calculation. b. Graph the probability of lot acceptance as a function of p. This is called an operating characteristic curve. They key is to store your lists of ps and your list of probabilities as vectors in R, and then to use these as the X and Y axes in a geom_line) plot. The code below should get you started. ps <-c() #Fill in the values of p, separated by commas probs <-c() #Fill in the acceptance probabilities, separated by commas ggplot (data=NULL, aes(x=ps, y=probs)) + geom_line() probs ps C. A quality control engineer is considering two different lot acceptance sampling plans: Plan 1: n=5, a=1 Plan 2: n=100, a=20. Construct operating characteristic curves for both sampling plans. You can make separate plots. Hint: use the normal approximation to a binomial to do calculations for the second plan d. If you were a seller producing lots with proportions of defectives between 0 and 10%, which plan would you prefer? Plan 1 or plan 2? Why? Write answer here then justify e. If you were a buyer wishing high protection against accepting lots with proportion of defectives over 30%, which plan would you prefer? Plan 1 or plan 2? Why? Write answer here and then justify

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Sophia 5 days 2021-10-08T02:04:24+00:00 1 Answer 0

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    2021-10-08T02:06:10+00:00

    Answer:

    Step-by-step explanation

    Let β = np, n= 5 and p= 0, 0.1, 0.3, 0.5, 0.9 and 1

    x > a, a= 0

    The probability of an item inspected, P(x) = β∧x*e∧-β/x!

    a) For x= 1: β = 5 X 0 = 0

    P(x=1) = 0

    For x = 2: β = 5 X 0.1 = 0.5

    P(x=2) = 0.5²*e∧-0.5/2!

    For x = 3: β = 5 X 0.3 = 1.5

    P(x=3) = 1.5³*e∧-1.5/3!

    For x =4: β = 5 X 0.5 = 2.5

    P(x=4) = 2.5∧4*e∧-2.5/4!

    For x = 5: β = 5 X 0.9 = 4.5

    P(x=5) = 4.5∧5*e∧-4.5/5!

    For x= 6: β = 5 X 1 = 5

    P(x= 6) = 5∧6*e∧-5/6!

    b) For plan1, x = 1: β + 10 X 0.1 = 1

    P(x=1) = e∧-1/1! = 1/e = 1/2.718 = 0.368

    For plan2, x=20: β = 100 X 0.1 = 10

    P(x=20) = 10∧20*e∧-10/20!

    Plan 2 is preferable because the probability of success is very small and the number of trial is comparatively large, hence, the poisson distribution will be suitable probability model.

    d) Plan 1 is prefered instead, because the probability of success is larger and more profit is made.

    e) Plen 2 is prefered, because it will be cheaper to afford.

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