Respuesta :
Answer:
False
Step-by-step explanation:
We are given the following information:
We treat adult who prefer one child to be a boy as a success.
P(prefer one child to be a boy) = 40% = 0.4
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15 and x = 8
We have to evaluate:
[tex]P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131[/tex]
Since the probability of 8 or more is 0.2131 is not very small, thus, it is not a rare event.
Thus, the given statement is false.