# Thread: Calculating Margin of Error Probability

1. ## Calculating Margin of Error Probability

Here is the question:

The monthly expenditures on food by single adults in Missoula are normally distributed with a mean of $410 and a SD of$70. What is the probability that the margin of error made in estimating the mean monthly expenditure of all single adults in that city by the mean of a random sample of 90 such adults will be at most $10? What I did: I tried using this equation, E=Z(a/2)SD/SQRT(N) => and get Z(a/2)=1.3553 which gives me a 91.15%. This seems really high though and I'm not sure what to do from here. Thanks! 2. Originally Posted by crabchef Here is the question: The monthly expenditures on food by single adults in Missoula are normally distributed with a mean of$410 and a SD of $70. What is the probability that the margin of error made in estimating the mean monthly expenditure of all single adults in that city by the mean of a random sample of 90 such adults will be at most$10?

What I did:

I tried using this equation, E=Z(a/2)SD/SQRT(N) => and get Z(a/2)=1.3553 which gives me a 91.15%. This seems really high though and I'm not sure what to do from here.

Thanks!
Solve $z_{\alpha/2} \frac{70}{\sqrt{90}} = 10$ for $z_{\alpha/2}$ (since it's two-sided) and hence for $\alpha$:

$z_{\alpha/2} = 1.3553 \Rightarrow \alpha = 2 (0.087667) = 0.18$.

So the required probability is $1 - \alpha = 0.82$.