Question

exercise 14-10 letter B.

i calculated with a final result 0.0004 but its being marked wrong. Please help understand.

Unit 3, Topic 14, Exercice 14-10 Candy Bar Welghts 0.015 1125 Suppose the wrapper of a certsin candy bar lsts its weight as 2.13 ounces. Naturally, the weights of individual bars vary somevhat. Suppose also that the weights of these candy bars vary according to a normal distribution with mean μ-2.2 ounces and standard deviation σ 0.04 ounces. a. What does the CLT say about the distribution of sample mean weights if samples of size n are taken repeatedly? Round your answer for the standard deviation to four decimal places. The sampling distribution of the mean weights will be normally distributed with mearn 7 ounces and standard deviation 2 ounces Suppose that you take a random sample of n 7 candy bars. b. Determine the probability that the sample mean weight would be less than 2. Round your answer to four decimal places. 0.0004 ounces. 3107 c. Determine the probability that the sample mean weight would be less than 2.19 Round your answer to four decimal places. -o.ou 035o ounces. a. ,9-a.3 -o.ol ·0.0,5:-aGG7 d. Determine the value such thathe robabliry 0.025 ad organing of obtaining a sample mean weight of less than that value. Round your answer to three decimal places. in o.04 行 ounces Answer 1: exact number, no tolerance Answer *2: the tolerance is +/-2% Answer * 3: the tolerance is +/-2% Answer * 4: the tolerance is +/-296 Answer * 5: the tolerance is +/-2% Question Attempts: O of 3 used

Answer 1

centsal timit Theorem som a Psobability distaibutio with 62u ane finite Then the Psobability bution of nado- Given n=7 x =213 2.13-2 o.04 o, 04 12-64S7; O. O 2 o.OIS

Date Page 2.13-2.2 X-시 2.13~2.2

Answer 2

[REMOVED]