I need help solving this question:

Question

1. In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 51 acres. Suppose we randomly survey 45 farmers from 1940.

I need help with the question below:

1. The middle 50% of the distribution for x̅, the bounds of which form the distance represented by the IQR, lies between what two values?

please explain to me the steps I have to do to solve this problem, thank you

Answer 1

You can solve this in Excel by using the functions below:

Lower = NORM.INV(.25,174,51)

Upper = NORM.INV(.75,174,51)

Here's the explanation:

We know that on the normal distribution curve, 50% falls to the left of the mean and 50% to the right.

You could picture the probabilities increasing as shown below moving from left to right

Similarly, the Z scores are shown underneath. At 50% which is the mean, Z = 0. We're interested in the Z values at 25% and 75%.

Probabilities--> 0% 25% 50% 75% 100%

Z score --> ? 0 ?

If we're looking for the middle 50% around the mean, that has to be what's between p = 25% and p = 75% (75%-25% = 50%).The NORM.INV function tells you exactly that.

Alternatively, you could look on a Z table for 25%, and 75% to get Z scores for each of them.

When you get the Z scores you could use the formula for Z scores:

Z = (x - mean) / stdev

If you substitute in the Z you found, the mean, and stdev, then all that's left to solve for is x (do this with both z scores). Hope this makes sense