Question

In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

The middle 50% of the distribution for X,the bounds of which form the distance represented by the IQR, lies between what two values? (Round your answers to two decimal places.)

Answer 1

The middle 50% of the distribution for X lies between 91.5 acres and 256.5 acres.

Explanation:

To find the middle 50% of the distribution for X, we need to calculate the values that represent the distance represented by the interquartile range (IQR).

First, let's calculate the IQR. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1).

Since the middle 50% of the data lies between Q1 and Q3, we can find the values for Q1 and Q3 using the standard deviation (SD) and the mean (174 acres).

Q1 = mean - (1.5 * SD)

Q3 = mean + (1.5 * SD)

Substituting the values:

Q1 = 174 - (1.5 * 55)

Q3 = 174 + (1.5 * 55)

Calculating:

Q1 = 174 - 82.5 = 91.5

Q3 = 174 + 82.5 = 256.5

So, the middle 50% of the distribution for X lies between 91.5 acres and 256.5 acres.