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 range within \(k = 2\) standard deviations of the mean in a normal distribution (empirical rule approximation for binomial when \(np\geq5\) and \(n(1 - p)\geq5\), here \(n = 4\), \(p=0.11\) (since \(11\%\) of employees blame oversleeping), \(\mu=np = 4\times0.11 = 0.44\), \(\sigma=\sqrt{np(1 - p)}=\sqrt{4\times0.11\times(1 - 0.11)}=\sqrt{4\times0.11\times0.89}\approx0.63\)). The range is \(\mu\pm k\sigma\).

The difference from the mean is \(k\sigma\). When \(k = 2\) (a common - used multiple for a reasonable range in non - strict binomial - to - normal approximation), we calculate \(2\sigma\).

Step2: Calculate \(2\sigma\)

Given \(\sigma = 0.63\), then \(2\sigma=2\times0.63 = 1.26\).

Answer:

\(1.26\)