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) support = female B. P(BC) Complement = Silver Preferred i. Male P(A)/ speak to = 60/ one cytosine thirty = .4615 ii. Silver P(BC)/ extreme = 80/ one hundred thirty = .6153 iii. P(A U B) = P(A) + P(B) P(A ? B) = 60/ one hundred thirty + 50/130 40/130 = .5384 iv. Male P(A)/ perfect = 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 atomic number 18 various they are not independent. 3. i. ii. No I entrust no debauch the contract because it exceeds the amount of the expect value iii. I lead be pass oning to pay less(prenominal) or agree to $7,700 4. i. P(x)= person leaves within a year. N=8 P=.40 PDF.Binom(0,8,.4) = .1679 portion of no one leaving. ii. P(x at least 4) 1-CDF.Binom(3,8,.4) = .4059 chance that at least 4 volition leave. iii.

p( amidst 5 and 7) =CDF.Binom(7,8,.4) CDF.Binom(4,8,.4) = .9993 - .8263 = .1730 hazard that between 5 and 7 comprehensive go forth leave. iv. P(Less than 3 Leave) = CDF.Binom(2,8,.06) = .3153 Probability that less than 3 will leave. v. P(x all 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 want to nonplus a full essay, order it on our website: Ordercustompaper.com

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