When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let X = the num

Question

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05). (Round your probabilities to three decimal places.)(a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.)

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Rose 4 days 2021-10-11T14:35:12+00:00 1 Answer 0

Answers ( )

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    2021-10-11T14:36:46+00:00

    Answer:

    a) P(X ≤ 2) = 0.87

    b)  P(X ≥ 5) = 0.01

    c) P(1 ≤ X ≤ 4) = 0.71

    d) P ( X = 0 ) = 0.28

    e) σ(X) = 1.09 , E(X) = 1.25

    Step-by-step explanation:

    Given:

    – Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.05)

    Where, n = 25 and p = 0.05

    Find:

    (a) Determine P(X ≤ 2).(b) Determine P(X ≥ 5).(c) Determine P(1 ≤ X ≤ 4).(d) What is the probability that none of the 25 boards is defective?(e) Calculate the expected value and standard deviation of X.

    Solution:

    – The probability mass function for a binomial distribution is given by:

                             P ( X = x ) = nCr * (p)^r * ( 1 – p )^(n-r)

    a) P(X ≤ 2):

                             P(X ≤ 2) = P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 )

                             = (0.95)^25 + 25*0.05*0.95^24 + 25C2*0.05^2*0.95^23

                             = 0.87

    b) P(X ≥ 5):

                P(X ≥ 5) = 1 – [P ( X = 0 ) + P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)

                = 1 – [ 0.87 + 25C3*0.05^3*0.95^22 + 25C4*0.05^4*0.95^21]

                = 1 – 0.98994

                = 0.01

    c) P(1 ≤ X ≤ 4):

                P(1 ≤ X ≤ 4) = P ( X = 1 ) + P ( X = 2 ) + P ( X = 3 ) + P(X=4)

                = ( 0.87 – 0.95^25) + 0.11994

               = 0.71

    d) P( X = 0 )

                P ( X = 0 ) = 0.95^25 = 0.28

    e) E(X) & σ(X):

                E(X) = n*p

                E(X) = 25*0.05 = 1.25

                σ(X) = sqrt ( Var (X) )

                σ(X) = sqrt ( n*p*(1-p) ) = sqrt ( 25*0.05*0.95 )

                σ(X) = 1.09

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