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4 November, 15:18

A random sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manufactured is 0.03. what is probability that the estimated probability of a nonconforming product will be less than or equal to 0.06 if the fraction of nonconforming really is 0.03

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  1. 4 November, 16:29
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    solution:

    The probability mass function for binomial distribution is,

    Where,

    X=0,1,2,3, ...; q=1-p

    find the probability that (p∧ ≤ 0.06), substitute the values of sample units (n), and the probability of nonconformities (p) in the probability mass function of binomial distribution.

    Consider x to be the number of non-conformities. It follows a binomial distribution with n being 50 and p being 0.03. That is,

    binomial (50,0.02)

    Also, the estimate of the true probability is,

    p∧ = x/50

    The probability mass function for binomial distribution is,

    Where,

    X=0,1,2,3, ...; q=1-p

    The calculation is obtained as

    P (p^ ≤ 0.06) = p (x/20 ≤ 0.06)

    = 50cx ₓ (0.03) x ₓ (1-0.03) 50-x

    = (50c0 ₓ (0.03) 0 ₓ (1-0.03) 50-0 + 50c1 (0.03) 1 ₓ (1-0.03) 50-1 + 50c2 ₓ (0.03) 2 ₓ (1-0.03) 50-2 + 50c3 ₓ (0.03) 3 ₓ (1 - 0.03) 50-3)

    = (ₓ (0.03) 0 ₓ (1-0.03) 50-0 + ₓ (0.03) 1 ₓ (1-0.03) 50-1 + ₓ (0.03) 2 ₓ (1-0.03) 50-2 ₓ (0.03) 3 ₓ (1-0.03) 50-3)
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