when circuit boards used in the manufacture of compact disc players are tested,
ID: 3327850 • Letter: W
Question
when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 10%. Let X = the number of defective boards in a random sample of size n = 25, so X~ Bin(25, 0.1). (Round your probabilities to three decimal places.) (a) Determine P(X 2) (b) Determine P(X 2 5). (c) Determine P(1 s X s 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.) expected value standard deviation boards boards You may need to use the appropriate table in the Appendix of Tables to answer this question.Explanation / Answer
p = 0.1
n = 25
P(X = x) = 25Cx * 0.1x * (1 - 0.1)25-x
a) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 25C0 * 0.10 * 0.925 + 25C1 * 0.11 * 0.919 + 25C2 * 0.12 * 0.918
= 0.537
b) P(X > 5) = 1 - P(X < 5)
= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)]
= 1 - [ 25C0 * 0.10 * 0.925 + 25C1 * 0.11 * 0.919 + 25C2 * 0.12 * 0.918 + 25C3 * 0.13 * 0.917 + 25C4 * 0.14 * 0.916 ]
= 1 - 0.902
= 0.098
c) P(1 < X < 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 25C1 * 0.11 * 0.919 + 25C2 * 0.12 * 0.918 + 25C3 * 0.13 * 0.917 + 25C4 * 0.14 * 0.916
= 0.830
d) P(X = 0) = 25C0 * 0.10 * 0.925 = 0.072
e) Expected value = n * p = 25 * 0.1 = 2.5
Standard deviation = sqrt(n * p * (1 - p)) = sqrt(25 * 0.1 * 0.9) = 1.5
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.