A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require a

Question

A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require an adjustment. What is the probability that in a sample of 20 window frames?

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Kaylee 3 weeks 2021-11-19T05:20:28+00:00 1 Answer 0 views 0

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    2021-11-19T05:22:04+00:00

    Answer:

    a) 0.3585

    b) 0.6415

    c) 0.07548

    Step-by-step explanation:

    This is a binomial distribution problem

    Binomial distribution function is represented by

    P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

    n = total number of sample spaces = 20

    x = Number of successes required

    p = probability of window frame with minor defect = 0.05

    q = probability of window frame with no defect = 1 – 0.05 = 0.95

    a) Probability that none of the frames picked will have defects

    x = 0

    P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

    P(X = 0) = ²⁰C₀ 0.05⁰ 0.95²⁰⁻⁰ = 0.3585

    b) Probability that At least one will need adjustment = P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0) = 1 – 0.3585 = 0.6415

    c) Probability that More than two will need adjustment = P(X > 2)

    P(X > 2) = 1 – P(X ≤ 2) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

    So, we evaluate each of that using the binomial distribution formula, slotting in the appropriate variables, and sum it all up.

    P(X > 2) = 1 – 0.9245 = 0.07548

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