Question

seventeen percent of u.s. employees who are late for work blame oversleeping. you randomly select four u.s. employees who are late for work and ask them whether they blame oversleeping. the random variable represents the number of u.s. employees who are late for work and blame oversleeping.

find the mean of the binomial distribution.
\\(\mu = 0.68\\) (round to the nearest hundredth as needed.)

find the variance of the binomial distribution.
\\(\sigma^2 = 0.56\\) (round to the nearest hundredth as needed.)

find the standard deviation of the binomial distribution.
\\(\sigma = 0.75\\) (round to the nearest hundredth as needed.)

interpret the standard deviation in the context of the real-life situation.
in most samples of four u.s. employees who are late for work, the number of u.s. employees that blame oversleeping would differ from the mean by no more than . values further from the mean than that would be considered unusual.
(type an integer or decimal rounded to the nearest hundredth as needed.)

Explanation:

Identify binomial parameters

$$
LATEXBLOCK0
$$

Calculate standard deviation

$$
LATEXBLOCK1
$$

Determine unusual value threshold

$$
LATEXBLOCK2
$$

Answer:

In most samples of four U.S. employees who are late for work, the number of U.S. employees that blame oversleeping would differ from the mean by no more than <blank>1.50</blank>. Values further from the mean than that would be considered unusual.