Question: Test: Module 3 Test Question 2 of 12 . : . 'bl This test 100 pomt(s) possl e Submit test This question: 14 point(s) possible

 Test: Module 3 Test Question 2 of 12 . : .

'bl This test 100 pomt(s) possl e Submit test This question: 14

Test: Module 3 Test Question 2 of 12 . : . 'bl This test 100 pomt(s) possl e Submit test This question: 14 point(s) possible The weight of an energy bar is approximately normally distributed with a mean of 42.40 grams with a standard deviation of 0.035 gram. Complete parts (a) through (e) below. f ..... \\ a. What is the probability that an individual energy bar weighs less than 42.375 grams? (Round to three decimal places as needed.) b. If a sample of 4 energy bars is selected, what is the probability that the sample mean weight is less than 42.375 grams? (Round to three decimal places as needed.) c. If a sample of 25 energy bars is selected, what is the probability that the sample mean weight is less than 42.375 grams? (Round to three decimal places as needed.) d. Explain the difference in the results of (a) and (c). Part (a) refers to an individual bar, which can be thought of as a sample with sample size . Therefore, the standard error of the mean for an individual bar is I: times the standard error of the sample in (c) with sample size 25. This leads to a probability in part (a) that is V the probability in part (c). (Type integers or decimals. Do not round.) 9. Explain the difference in the results of (b) and (c). The sample size in (c) is greater than the sample size in (b), so the standard error of the mean (or the standard deviation of the sampling distribution) in (c) is V than in (b). As the standard error V values become more concentrated around the mean. Therefore, the probability that the sample mean will fall close to the population mean will always V when the sample size increases

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