Question

3. According to a survey, 68% of Americans say that
they go to a movie theater at least once a month.
Suppose that this proportion is true for the current
American population.
a.
Let x be a binomial random variable that denotes
the number of Americans in a random sample of
12 who say that they go to a movie theater at
least once a month. What are the possible
values that x can assume?
b. Find the probability that exactly 9 Americans in
a random sample of 12 will say that they go to a
movie theater at least once a month.

Answer 1

Answer:

The probability that exactly 9 Americans in  a random sample of 12 will say that they go to a  movie theater at least once a month

P( X = 9) =  0.007343

Step-by-step explanation:

Step(i):-

According to a survey, 68% of Americans say that  they go to a movie theater at least once a month.

Given proportion 'p' = 68% =0.68

                             q = 1-p = 1- 0.68 = 0.32

Given Number of Americans in a random sample

'n' = 12

Let 'X' be random variable of binomial distribution

$P( X = r) = n_{C_{r} } p^{r} q^{n-r}$

Step(ii):-

Given r = 4

Given random sample 'n' =12

The probability that exactly 9 Americans in  a random sample of 12 will say that they go to a  movie theater at least once a month

$P( X=9) = 12_{C_{9} }( 0.68)^{9} (0.32)^{12-9}$

using factorial notation

      $12 C_{9} = 12_{C_{12-9} } = 12_{C_{3} } = \frac{12!}{(12-3)!3!} = \frac{12 X 11 X 10 X 9!}{9!3 X 2 X 1} = \frac{12 X 11 X 10}{3 X 2 X 1} = 220$

$P( X=9) = 220( 0.68)^{9} (0.32)^{12-9}$

On calculation , we get

P( X = 9) =  0.007343

Final answer:-

The probability that exactly 9 Americans in  a random sample of 12 will say that they go to a  movie theater at least once a month

P( X = 9) =  0.007343