Question

eleven 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.
μ = .44 (round to the nearest hundredth as needed.)
find the variance of the binomial distribution.
σ² = .39 (round to the nearest hundredth as needed.)
find the standard deviation of the binomial distribution.
σ = .63 (round to the nearest hundredth as needed.)
interpret the results in the context of the real - life situation.
in most samples of four adults are late for work, the average number of adults that blame oversleeping would differ from the mean by no more than □.
(type an integer or decimal rounded to the nearest hundredth as needed.)

Explanation:

Step1: Recall the formula for the standard deviation of a binomial distribution

The standard deviation \(\sigma\) of a binomial distribution is given by \(\sigma=\sqrt{np(1 - p)}\), where \(n\) is the number of trials and \(p\) is the probability of success. Here, \(n = 4\) and \(p=0.11\) (since \(11\%=0.11\)), and we already know \(\sigma = 0.63\) (from the problem statement). The question is asking for the value within \(1\) standard deviation of the mean.

Step2: Calculate the range within \(1\) standard deviation of the mean

The range within \(1\) standard deviation of the mean is \(\mu-\sigma\) to \(\mu+\sigma\). We know \(\mu = 0.44\) and \(\sigma=0.63\).

• Lower bound: \(0.44 - 0.63=-0.19\) (but since the number of people can't be negative, we consider \(0\) as the lower - bound in the context of counting people).
• Upper bound: \(0.44 + 0.63 = 1.07\approx1\) (since we are counting the number of people, we round to the nearest whole number).

Answer:

\(1\)