Asked by Rhodey
The mean per capita income is 24,787 dollars per annum with a variance of 169,744. What is the probability that the sample mean would differ from the true mean by less than 42 dollars if a sample of 412 persons is randomly selected?
I know what formula to use for this question (z: x-μ/(σ/√n)) and I know that the standard deviation would be σ= √169 744 = 142. But, I'm just confused about what this statement means: "probability that the sample mean would differ from the true mean by less than 42 dollars"?
I know what formula to use for this question (z: x-μ/(σ/√n)) and I know that the standard deviation would be σ= √169 744 = 142. But, I'm just confused about what this statement means: "probability that the sample mean would differ from the true mean by less than 42 dollars"?
Answers
Answered by
PsyDAG
Z = (24,745-24,787)/(142/20) = ?
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Since you want to know above AND below the mean, double that value.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Since you want to know above AND below the mean, double that value.
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