A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the standard deviation of weights is 5.5 ounces. (This standard deviation is therefore known.) A quality control expert selects n containers at random. How large a sample would be required in order for the 99% confidence interval for to have a length of 2 ounces? a. n 15 b. n 16 c. n 201 d. n 226
Accepted Solution
A:
Answer:option (c) n = 201Step-by-step explanation:Data provided in the question:Standard deviation, s = 5.5 ounceConfidence level = 99%Length of confidence interval = 2 ouncesTherefore, margin of error, E = (Length of confidence interval ) ÷ 2= 2 ÷ 2= 1 ounceNow,E = [tex]\frac{zs}{\sqrt n}[/tex]here, z = 2.58 for 99% confidence intervaln = sample sizethus,1 = [tex]\frac{2.58\times5.5}{\sqrt n}[/tex]orn = (2.58 × 5.5)²orn = 201.3561 ≈ 201Hence,option (c) n = 201