## 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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2021-10-11T14:35:12+00:00
2021-10-11T14:35:12+00:00 1 Answer
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## Answers ( )

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

= 0.87b) 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

= 0.01c)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

= 0.71d) P( X = 0 )

P ( X = 0 )= 0.95^25 =0.28e) E(X) & σ(X):

E(X) = n*p

E(X)= 25*0.05 =1.25σ(X) = sqrt ( Var (X) )

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

σ(X) = 1.09