Answer:
the probability that more than 9 but fewer than 14 disapprove of smoking pot daily is 0.608
Step-by-step explanation:
From the information given:
Let N represent the total number of senior students
N = 17000
and k to be the number of seniors disapproving of daily pot smoking
k = 17000 × 0.70
k = 11900
sample size n = 18
Suppose X represents the random variable...
Then,
X = 0,1,2,...,18
Thus X follows a hypergeometric distribution with parameters
(N = 17000, n = 18 & k = 11900)
Thus, the probability mass function is:
[tex]P(X =x) = \dfrac{(^k_n) (^{N-k}_{n-x} ) }{(^N_n) }; \ max \{0,(N-K) \} \le x \le min \{n,k\}[/tex]
[tex]P(X =x) = \dfrac{(^{11900}_n) (^{17000-11900}_{18-x} ) }{(^{17000}_{18}) }[/tex]
[tex]P(X =x) = \dfrac{(^{11900}_n) (^{51000}_{18-x} ) }{(^{17000}_{18}) }; \ 0\le x \le 18[/tex]
Now, the required probability is computed as:
[tex]P(9<X<14) = P(10 \le X \le13)[/tex]
[tex]P(9<X<14) = P(X =10) +P(X =11) +P(X=12) +P(X =13)[/tex]
[tex]P(9<X<14) = \left\{\begin{array}{c} \dfrac{(^{11900}_{10}) (^{5100}_{18-10}) }{(^{17000}_{18})} + \dfrac{(^{11900}_{11}) (^{5100}_{18-11}) }{(^{17000}_{18})}+ \dfrac{(^{11900}_{12}) (^{5100}_{18-12}) }{(^{17000}_{18})} + \dfrac{(^{11900}_{13}) (^{5100}_{18-13}) }{(^{17000}_{18})}\end{array}\right\}[/tex]P(9<X<14) = 0.0811 + 0.1377 + 0.1874 + 0.2018
P(9<X<14) = 0.608
Thus, the probability that more than 9 but fewer than 14 disapprove of smoking pot daily is 0.608
