Thank you in advance When circuit boards used in the manufac

Thank you in advance

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 n 25, 0.1). (Round your probabilities to three decimal places.) (a) Determine p(x S 2 53.709 x (b) Determine PX z 5). 86.158 x (c) Determine PO1 s X S 4) 83.027 (d) What is the probability that none of the 25 boards is defective? 7.1789 (e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.) boards expected value boards standard deviation

Solution

P(X = 0) = (25C0) * (0.9)^25 * (0. 1)^0 = 0.07179

P(X = 1) = (25C1) * (0.9)^24 * (0. 1)^1 = 0.19942

P(X = 2) = (25C2) * (0.9)^23 * (0. 1)^2 = 0.26589

P(X = 3) = (25C3) * (0.9)^22 * (0. 1)^3 = 0.22650

P(X = 4) = (25C4) * (0.9)^21 * (0.1)^4 = 0.13842

P(X = 5) = (25C5) * (0.9)^20 * (0.1)^5 = 0.06459

a) P(X <= 2) = P(X = 2) + P(X = 1) + P(X = 0)

= 0.26589 + 0.19942 + 0.07179

= 0.53709 = 53.709%

b) P(X >= 5) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3) - P(X = 4) - P(X = 5)

= 1 - 0.96660

= 0.03340= 3.340%

c) P(1 <= X <= 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.83022 = 83.022%

d) P(X = 0) = 0.07179 = 7.179%

e) (i) E(X) = n*p = 25*0.10 = 2.5 defectives

(ii) SD(X) = (n*p (1-p))^0.5 = (25*0.10*0.90)^0.5

= (2.25)^0.5 = 1.5