Question

Binomial Distribution. Surveys repeatedly show that about 40% of adults in the U.S. indicate that if they only had one child, they would prefer it to be a boy. Suppose we took a random sample of 15 adults and the number who indicated they preferred a boy was 8. This would be considered a rare event because the probability of 8 or more is so low.
True/False

Answer 1

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

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

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:

$P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131$

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.