In sampling a large number of parts are manufactured by a machine, the mean number of defective in a sample of 20 is 2,out of 1000 such samp

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In sampling a large number of parts are manufactured by a machine, the mean number of defective in a sample of 20 is 2,out of 1000 such samples. How many would br expected to contain atleast 3 defective parts?

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Genesis 1 month 2021-09-21T23:26:14+00:00 1 Answer 0 views 0

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    2021-09-21T23:28:02+00:00

    Given : In sampling a large number of parts are manufactured by a machine, the mean number of defective in a sample of 20 is 2,out of 1000 such samples. How many would br expected to contain atleast 3 defective parts ?

    Solution :

    As given , this is case of simple binomial. we are given that probability that an item would be defective in a sample of 20 is 2 .

    So,

    → P = (2/20) = (1/10) = 0.1

    As, we know,

    → Q = 1 – P

    So,

    → Q = 1 – 0.1 = 0.9 . { Probability of non defective .}

    now we have , n = 20 and N = 1000 .

    as we know,

    • P(x = X) = n(C)x * P^x * q^(n – x)

    So, in order to find 3 expected defective :-

    → P(x = 3) = 20(C)3 * (0.1)³ * (0.9)¹⁷ = 0.323 .

    Therefore,

    → The number of samples having at least three defective parts out of 1000 samples = 1000 * 0.323 = 323. (Ans.) .

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