Answer:
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
