Question

A recent survey of 1010 employees found that 25% of them would lay off their bosses if they could. Complete parts a through d below based on a random sample of 10 employees. a. What is the probability that exactly three employees would lay off their boss? The probability is 0.2503 . (Round to four decimal places as needed.) b. What is the probability that three or fewer employees would lay off their bosses? The probability is 0.7759. (Round to four decimal places as needed.) c. What is the probability that five or more employees would lay off their bosses? The probability is 0.0781 . (Round to four decimal places as needed.) d. What are the mean and standard deviation for this distribution? The mean number of employees that would lay off their bosses is (Type an integer or a decimal. Do not round.) .

Answer 1

solution

q = 1 - 0.25 = 0.75

Mean = $\mu$ = n * p = 10 * 0.25= 2.5

Standard deviation = $\sigma$ = $\sqrt$ n * p * q = $\sqrt$ 10 * 0.25 * 0.75 = 1.3693