Questions: 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= (Round to the nearest hundredth as needed.)

Solution Steps

We are given the following parameters for the binomial distribution:

• Number of trials \( n = 4 \)
• Probability of success \( p = 0.11 \)
• Probability of failure \( q = 1 - p = 0.89 \)

The mean \( \mu \) of a binomial distribution can be calculated using the formula: \[ \mu = n \cdot p \] Substituting the values: \[ \mu = 4 \cdot 0.11 = 0.44 \]

The variance \( \sigma^2 \) of a binomial distribution is given by: \[ \sigma^2 = n \cdot p \cdot q \] Substituting the values: \[ \sigma^2 = 4 \cdot 0.11 \cdot 0.89 \approx 0.39 \]

The standard deviation \( \sigma \) is the square root of the variance: \[ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{0.39} \approx 0.63 \]

Final Answer