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.
μ = 4.2 (round to two decimal places as needed.)
find the variance of the binomial distribution.
σ² = 0.67 (round to two decimal places as needed.)
find the standard deviation of the binomial distribution.
σ = 0.82 (round to two decimal places as needed.)
interpret the results.
on average, ▢ out of every 5 adults in the country believe that life on other planets is plausible. in most samples of 5 adults, the number of adults who think life on other planets is plausible would differ from the mean by no more than ▢.
(type integers or decimals rounded to two decimal places as needed.)

Explanation:

Step1: Interpret the results

The mean $\mu = 4.2$ means that on average, out of every 5 adults, 4.2 believe that life on other planets is plausible.

Step2: Find the variance

Given $\sigma^{2}=0.67$. Variance measures the spread of the data points around the mean.

Step3: Find the standard deviation

Given $\sigma = 0.82$. Standard deviation is the square - root of the variance. It also measures the spread. The formula for standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$, and the formula for variance is $\sigma^{2}=np(1 - p)$ where $n = 5$ (number of trials, i.e., number of adults selected) and $p=0.84$ (since 84% or 0.84 of adults believe).

Answer:

On average, out of every 5 adults, 4.2 believe that life on other planets is plausible. The variance $\sigma^{2}=0.67$ and standard deviation $\sigma = 0.82$ measure the spread of the number of adults (in samples of 5) who believe life on other planets is plausible around the mean value of 4.2.