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 = □ (round to the nearest hundredth 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 \(n = 4\) (number of trials, i.e., number of employees selected) and \(p=0.11\) (probability that an employee is late for work and blames oversleeping).

Step2: Calculate the variance

We know that \(\sigma^{2}=np(1 - p)\). Substitute \(n = 4\) and \(p = 0.11\) into the formula:

$$
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Step3: Calculate the standard deviation

Since \(\sigma=\sqrt{\sigma^{2}}\), and \(\sigma^{2}=0.3916\) (from the previous step)

$$ \sigma=\sqrt{0.3916}\approx0.63 $$

Answer:

\(\sigma\approx0.63\)