SOLUTION: . One in every three Americans believes that the U.S government should send a mission to Mars. If n = 30 Americans are randomly selected, find the probability that Part a. Exactly

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Question 1198375: . One in every three Americans believes that the U.S government should send a
mission to Mars. If n = 30 Americans are randomly selected, find the probability that Part a. Exactly 8 out of these 30 selected Americans believe that the U.S government should send a mission to Mars.
Part b. At least 8 out of these 30 selected Americans believe that the U.S government should send a mission to Mars No need to show calculations - you can use software)
Part c. Use the Normal Distribution to find the approximate answer in Part b.
Answer by ikleyn(52852) About Me  (Show Source):
You can put this solution on YOUR website!
.
One in every three Americans believes that the U.S government should send a
mission to Mars. If n = 30 Americans are randomly selected, find the probability that
Part a. Exactly 8 out of these 30 selected Americans believe that the U.S government
should send a mission to Mars.
Part b. At least 8 out of these 30 selected Americans believe that the U.S government
should send a mission to Mars No need to show calculations - you can use software)
Part c. Use the Normal Distribution to find the approximate answer in Part b.
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(a)  It is a typical binomial distribution problem.

     The number of trials is n=30;  the number of success trials is k=8;\
     The probability of success for each individual trial is p=1/3.

     Using standard designations, the probability is  P = P(n=30; k=8; p=1/3).


     To facilitate my calculations, I used online calculator at this site  https://stattrek.com/online-calculator/binomial.aspx

     It provides nice instructions  and  a convenient input and output for all relevant options/cases.


     The resulting number is P = 0.1192   (rounded).    ANSWER



(b)  Use the same mantra.

     The resulting number is P = P(n=30; k>=8; p=1/3) = 0.83321   (rounded).    ANSWER



(c)  The normal distribution approximation has the mean m= n*p = 30*(1/3) = 10
     and standard deviation of  SD = sqrt%28p%2An%2A%281-p%29%29 = sqrt%28%281%2F3%29%2A30%2A%281-1%2F3%29%29 = 2.582 (rounded).

     
     Use the continuous correction factor to get 

         P = normalcfd(7.5, 9999, 10, 2.582) = 0.8335  (rounded).    ANSWER

     It is close enough to the value of 0.83321 from part (b).

Solved.