A sample of 100 bugs yields an average weight of 5.1 ounces. Assuming that the p

Question

A sample of 100 bugs yields an average weight of 5.1 ounces. Assuming that the population standard deviation is 0.2 ounces, find the margin of error in estimating mu (the true mean weight of all bugs of this type) at the 95 % level of confidence. 1.96 0.02 Selected Answer: C. 0.104 Correct Answer: B. 0.039 A paint manufacturer claims its paint will last an average of 8 years. Assume that paint life is normally distributed with a standard deviation of 4.1 years. Suppose a random sample of 15 houses are painted with this manufacturer's paint. What is the probability that the average life of the paint for this sample will be more than 11 years? (Choose the closest value) Selected Answer: B. 0.2322 Correct Answer: A. 0.0023

Explanation / Answer

Q1)

sample size = 100

sample average weight = 5.1

population standard deviation = 0.2

standard deviation for = standard deviation /n = 0.2/100 = 0.2/10 = 0.02

margin of error for at 95% level of confidence = 1.96*(standard deviation for ) = 1.96*0.02 = 0.039

Q2)

avearge = 8

standard deviation = 4.1

sample size = 15

standard deviation for average life = standard deviation/n = 4.1/15 = 1.0586

z value for 11 is (11-8)/1.0586 = 2.8339, correspoding p value using z table is 0.9977

P(average life<11) = 0.9977

P(average life>11) = 1-0.9977 = 0.0023

probability that the average life of the paint for this sample will be more than11 years = 0.0023

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