Thread: 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!

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 for (since it's two-sided) and hence for :

.

So the required probability is .