Data

1. P(gas)(a)= .9 P(Something Else)(b)= .2 P(Both)= P(a) *P(b)= .9 *.2= .18 2. P(A)= anthropoid P(B)= Black Preferred A. P(AC) complement = Female B. P(BC) Complement = Silver Preferred i. antheral P(A)/ hail = 60/ iodin hundred thirty = .4615 ii. Silver P(BC)/Total = 80/cxxx = .6153 iii. P(A U B) = P(A) + P(B) P(A ? B) = 60/130 + 50/130 40/130 = .5384 iv. Male P(A)/Total = 40/130= .3076 v. P(BC) if Male/Total Males= 20/60 = .3333 vi. P(B) if Female/Black P. = 10/50= 0.2 vii. P(A ? B) = P(A) * P(B) ; 40/130 = 60/130 * 50/130? 40/130= .3076, and P(A) * P(B) = .1775 Since the two results are motley they are not independent. 3. i. ii. No I entrust no buy the contract because it exceeds the amount of the expected value iii. I forget be willing to pay less or agree to $7,700 4. i. P(x)= Someone leaves within a year. N=8 P=.40 PDF.Binom(0,8,.4) = .1679 fortune of no one leaving. ii. P(x at to the lowest degree 4) 1-CDF.Binom(3,8,.4) = .4059 chance that at least 4 will leave. iii.

p(between 5 and 7) =CDF.Binom(7,8,.4) CDF.Binom(4,8,.4) = .9993 - .8263 = .1730 Probability that between 5 and 7 inclusive will leave. iv. P(Less than 3 Leave) = CDF.Binom(2,8,.06) = .3153 Probability that less than 3 will leave. v. P(x only 8 will leave) PDF.Binom(8,8,.4) = .0006 Probability that 8 will leave. (very low) vi. 5. a. 1-CDF.Binom(239,256,.95)= .8551 b. 1-CDF.Binom(250,258,.95) = .0525 c. .7682 - .1590 = .6090 d. 257 Tickets e. 259 TickIf you requir e to nonplus a full essay, order it on our ! website: BestEssayCheap.com

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