Question

Twenty-six 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. u= (Round to the nearest hundredth as needed.) Find the variance of the binomial distribution. o2 = N (Round to the nearest hundredth as needed.) Find the standard deviation of the binomial distribution. o= 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.)

Answer 1

DATE Ansa - p = 0.26 n=4 let, x= the number of U.S. employees who are late for work and blame overrsleeping. xabinomial (hp) mean - q) 4 = np = 4(0.26) = 11.04 b) Variance = = = np(1-P) 4 x 0.26 x(1-0.25) storia 0.77 c) standard deviation . - 8 = variance = 50.77 0.88 ] In most samples of four adults are late for work , the average number of adults that blame over sleeping would differ from the mean by no more than 10.88 Scanned with Camscanner