A poll reported that 65% of adults were satisfied with the job the major airlines were doing. Suppose 20 adults are selected at random and the number who are satisfied are recorded. Find the probability that exactly 11 adults are satisfied with the airlines.

the probability that 11 adults are satisfied is 0.115 (11.5%)

Step-by-step explanation:

since the result of each adult is independent of others , then since the adults are selected at random , the random variable X= number of adults who are satisfied , has a binomial probability distribution , where

P(X=x) = n!/[(n-x)!*x!] * p^x * (1-p)^(n-x)

where

n= sample size = 20

p= probability of any adult to be satisfied with the job of airlines = 0.65

x = number of adults who are satisfied

P(X=x) = probability that x adults are satisfied

then for x= 11 , we have

P(X=11) =   20!/[(20-11)!*11!] * 0.65^11 * (1-0.65)^(20-11) = 0.115 (11.5%)

then the probability that 11 adults are satisfied is 0.115 (11.5%)

answer: i think one of the answer is $150 and $300?

step-by-step explanation: