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.)

Answer 1

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