One of the products of a company is flour in 1000 gr packages. The weights of these packages actually follow a normal distribution with a known standard deviation, equal to 35 grams. On a certain day, a simple random sample of 125 containers has been selected, and its average weight has been measured to be 985 grams: [Value: 4 pts.] a) Calculate a 95.5% confidence interval for the average weight of the packages packed on the day. b) What would the previous confidence interval be if the confidence level were set at 99.7%? c) What is the probability that a package weighs less than 930 gr. ? d) What is the probability that a package weighs between 970 gr. and 1030 gr. ? e) What is the probability that a package weighs 1000 gr. ? f) What is the probability that a package weighs more than 1000 gr. ? g) How many packages weigh less than or equal to 980 grams? h) How many packages weigh between 985 and 1015 grams? i) How many packages weigh between 990 and 1020 grams? j) How many packages weigh more than 1000 grams?

Question

Answer:

a) To calculate a 95.5% confidence interval for the average weight of the packages, you can use the formula for the confidence interval for the population mean when the sample size is sufficiently large: Confidence Interval (CI) = x̄ ± Z * (σ / √n) Where: x̄ is the sample mean (985 grams). Z is the critical value for the desired confidence level. For a 95.5% confidence level, Z is approximately 1.96 (you can find this value in standard normal distribution tables). σ is the population standard deviation (35 grams). n is the sample size (125 packages). Plugging in the values: CI = 985 ± 1.96 * (35 / √125) CI ≈ 985 ± 6.14 So, the 95.5% confidence interval for the average weight is approximately (978.86, 991.14) grams.

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