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"?
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.
The statement "probability that the sample mean would differ from the true mean by less than 42 dollars" refers to the likelihood of randomly selecting a sample of 412 persons whose mean per capita income is within 42 dollars of the true mean per capita income.
In other words, we want to calculate the probability that the difference between the sample mean and the true mean is less than 42 dollars.
To calculate this probability, we can use the z-score formula you mentioned: z = (x - μ) / (σ / √n), where x is the sample mean, μ is the true mean, σ is the standard deviation, and n is the sample size.
In this case, we are given that the true mean per capita income is 24,787 dollars per annum and the standard deviation is 142 dollars. The sample size is 412 persons.
To calculate the probability, we need to find the z-score corresponding to a difference of 42 dollars from the true mean. We can then use a standard normal distribution table or statistical calculator to find the cumulative probability associated with that z-score.
The z-score formula becomes: z = (42) / (142 / √412)
Next, we can plug in the values and calculate the z-score.
z = (42) / (142 / √412) ≈ (42) / (142 / 20.297) ≈ 2.998
Using a standard normal distribution table or statistical calculator, we can find the cumulative probability associated with a z-score of 2.998. This will give us the probability that the sample mean would differ from the true mean by less than 42 dollars.
Please note that different resources may give slightly different probability values due to rounding and approximation methods.