Sixty-eight percent of us adults have little confidence in their cars. you randomly select eleven us adults. find the probability that the number of us adults who have little confidence in their cars is (1) exactly eight and then find the probability that it is (2) more than 6.

1) P(X=8) = 24.71%

2) P(X>6) = 74.37%

Step-by-step explanation:

since each experiment is individual and is not affected by the others, and also since the sample size is small enough compared with the population size, then X = number of adults that have little confidence in their cars follow a binomial distribution

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

where n= sample size=11 , p= probability of success for a single trial =68% x=observed number of adults that have little confidence in their cars

therefore

for a)

P(X=8) = 11!/[(11-8)!8!]*0.68^8*(1-0.68)^(11-8) = 0.2471 = 24.71%

for b)

P(X>6) = 1 - P(X≤6)

where

P(X≤6) = ∑ P(X≤i) for i from 0 to 6

P(X≤6) = 0.2563

therefore

P(X>6) = 1 - 0.2563 = 0.7437 = 74.37%

3x + 4y -25 =0 represents the amount of change sianna should recieve