A company is interested in estimating the mean number of days of sick leave taken by its employees. The firm's statistician randomly selects 100 personnel files and notes the number of sick days taken by each employee. The sample mean is 12.2 days and the sample standard deviation is 10 days. How many personnel files would the director have to select in order to estimate ? to within 2 days with a 99% confidence interval?(A) 2(B) 13(C) 136(D) 165(E) 166

Answer:E) 166Step-by-step explanation:Standard deviation is 10 Margin error for the problem is 2 Probability 99%, that means thet the siginficance level α is 1 – p α = 1 – 0.99 = 0.01 margin of error (ME) can be defined as follows ME = Z(α/2) * standard deviation/ √n Where n is the sample size Z(0.01/2) = Z(0.005) Using a z table Z = 2.576 Now, replacing in the equation and find n 2 = 2.576 * 10/ √n 2 = 25.76 /√n √n = 25.76 /2 √n = 12.88 n = 12.88^2 n = 165.89 ≈166