SOLUTION: Of those mountain climbers who attempt Mt. McKinley, only 65% reach the summit. In a random sample of 12 mountain climbers who are going to attempt Mt. McKinley, what is the probab

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Question 666404: Of those mountain climbers who attempt Mt. McKinley, only 65% reach the summit. In a random sample of 12 mountain climbers who are going to attempt Mt. McKinley, what is the probability of each of the following?
a.)At least 8 reach the summit
b.)no more than 8 reach the summit
c.)6 to 10 reach the summit, including 6 and 10
Answer by swincher4391(1107) (Show Source):
You can put this solution on YOUR website!
You either reach the summit or you don't. Thus we have a binomial distribution.
a) P[x>=8] = P[x=8]+P[x=9]+P[x=10]+P[x=11]+P[x=12] = (12 choose 8) * (.65)^8 * (.35)^4 + (12 choose 9) * (.65)^9 * (.35)^3 + (12 choose 10) * (.65)^10 * (.35)^2 + (12 choose 11) * (.65)^11 * (.35)^1 + (12 choose 12) * (.65)^12 * (.35)^0 = .5833
b) P[x<=8] = P[x<8] + P[x=8] = (1-P[x>=8]) + P[x=8]. As you see, we used the complement. We already know what P[x>=8] and its complement is P[x<8]. Saves us a lot of work.
(1-.5833) + (12 choose 8)*(.65)^8*(.35)^4 = .6534
c)P[6<=x<=10] = P[x=6]+P[x=7]+P[x=8]+P[x=9]+P[x=10]. Try to work this one out yourself. You should get .8730.