Suppose that 70% of all Americans agree that a candidate is not trustworthy. A survey finds the 116 out of 200 agree that a candidate is not trustworthy. How likely is the result above to happen, if the sample is representative of the population?

Prob = 0.00009 = 0.009%  , Less Likely

Step-by-step explanation:

Using Binomial Distribution, Prob = N c r . ( P )^ r . ( Q ) ^ (n -r)   .

,P = success probability, ie candidate considered be trustworthy = 70% = 0.7; N = trials = 200 ; Q = failure probability, ie non trustworthy = 1 - 0.7 = 0.3  r = no. of success = 116

Pr = 200c116 . (0.7)^116 . (0.3)^(200 - 116) = 200c116 (0.7)^116 (0.3)^84 = 0.00009 = 0.009%

Less probability means Less Likely