The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 51 households is monitored for one year to determine aluminum usage. If the population standard deviation of annual usage is 12.2 pounds, what is the probability that the sample mean will be each of the following? Appendix A Statistical Tables a. More than 61 pounds

0.007

Step-by-step explanation:

We were told in the above question that a random sample of 51 households is monitored for one year to determine aluminum usage

Step 1

We would have to find the sample standard deviation.

We use the formula = σ/√n

σ = 12.2 pounds

n = number of house holds = 51

= 12.2/√51

Sample Standard deviation = 1.7083417025.

Step 2

We find the z score for when the sample mean is more than 61

z-score formula is z = (x-μ)/σ

where:

x = raw score = 61 pounds

μ = the population mean = 56.8 pounds

σ = the sample standard deviation = 1.7083417025

z = (x-μ)/σ

z = (61 - 56.8)/ 1.7083417025

z = 2.45852

Finding the Probability using the z score table

P(z = 2.45852) = 0.99302

P(x>61) = 1 - P(z = 2.45852) = 0.0069755

≈ 0.007

Therefore,the probability that the sample mean will be more than 61 pounds is 0.007

3; 9 is 8 to -4

step-by-step explanation:

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