When circuit boards used in the manufacture of compact discplayers are tested, t

Q: When circuit boards used in the manufacture of compact discplayers are tested, t

a) For n=0 to 2, c(25,n)*.05^n*.95^(n-1)= b) For n=0 to 4, 1 - c(25,n)*.05^n*.95^(n-1)= c) For n=1 to 4, c(25,n)*.05^n*.95^(n-1)= d) c(25, 0)*.05^0*.95^25 = e) For n=0 to 25 x(c(25,n)*.05^n*.95^(n-1)) = expected value Standard deviation: For n=0 to 25 (x-expectedvalue)^2

ID: 2918586 • Letter: W

Question

When circuit boards used in the manufacture of compact discplayers are tested, the long-run percentage of defectives is 5%.Let X= the number of defective boards in a random sample of sizen=25, so X~Bin(25, .05). 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 isdefective? e.) Calculate the expected value and standard deviation ofX. When circuit boards used in the manufacture of compact discplayers are tested, the long-run percentage of defectives is 5%.Let X= the number of defective boards in a random sample of sizen=25, so X~Bin(25, .05). 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 isdefective? e.) Calculate the expected value and standard deviation ofX.

Explanation / Answer

a) For n=0 to 2, c(25,n)*.05^n*.95^(n-1)= b) For n=0 to 4, 1 - c(25,n)*.05^n*.95^(n-1)= c) For n=1 to 4, c(25,n)*.05^n*.95^(n-1)= d) c(25, 0)*.05^0*.95^25 = e) For n=0 to 25 x(c(25,n)*.05^n*.95^(n-1)) = expected value Standard deviation: For n=0 to 25 (x-expectedvalue)^2

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When circuit boards used in the manufacture of compact disc players are tested,

When circuit boards used in the manufacture of compact discplayers are tested, t

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When circuit boards used in the manufacture of compact discplayers are tested, t

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