Question

eighty - four percent of adults in a certain country believe that life on other planets is plausible. you randomly select five adults and ask them whether they believe that life on other planets is plausible. the random variable represents the number of adults who believe that life on other planets is plausible. find the mean, variance, and standard deviation of the binomial distribution for the random variable. interpret the results.
find the mean of the binomial distribution.
\\( \mu = 4.2 \\) (round to two decimal places as needed.)
find the variance of the binomial distribution.
\\( \sigma ^ { 2 } =.67 \\) (round to two decimal places as needed.)
find the standard deviation of the binomial distribution.
\\( \sigma = \square \\) (round to two decimal places as needed.)

Explanation:

Step1: Recall binomial distribution formulas

For a binomial distribution \(X\sim B(n,p)\), the mean \(\mu = np\), the variance \(\sigma^{2}=np(1 - p)\), and the standard deviation \(\sigma=\sqrt{np(1 - p)}\). Here \(n = 5\) (number of trials, i.e., number of adults selected) and \(p=0.84\) (probability of success, i.e., probability that an adult believes life on other planets is plausible).

Step2: Calculate the standard deviation

We know that \(\sigma^{2}=np(1 - p)=5\times0.84\times(1 - 0.84)\).
First, \(1-0.84 = 0.16\). Then \(5\times0.84\times0.16=5\times0.1344 = 0.67\) (variance \(\sigma^{2}\) is already given as \(0.67\)).
Now, \(\sigma=\sqrt{\sigma^{2}}=\sqrt{0.67}\). Using a calculator, \(\sqrt{0.67}\approx0.82\).

Answer:

\(0.82\)