Question

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 38 farmers from 1940.

a. In words, define the random variable X.

b. In words, define the random variable X-bar

c. Give the distribution of X-bar

(Round your standard deviation to two decimal places.)

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

smaller value _________ (acres)

larger value ________ (acres)

Answer 1

Solution:

We are given

µ = 174

σ = 51

n = 38

Part a

The random variable X is the size of the U.S. farms.

Part b

The random variable X-bar is the average size of the samples of 38 U.S.farms

Part c

The distribution of X-bar is given as below:

Mean for sampling distribution of X-bar = µ = 174

Standard deviation for sampling distribution of X-bar = σ/sqrt(n) = 51/sqrt(38) = 8.273292478

Standard deviation for sampling distribution of X-bar = 8.27

Part d

Z value for Q1 = -0.6745

Z value for Q3 = 0.6745

(by using z-table)

Q1 = µ + Z*σ/sqrt(n) = 174 - 0.6745*8.27 = 168.4219

Q3 = µ + Z*σ/sqrt(n) = 174 + 0.6745*8.27 = 179.5781

Smaller value = 168.42 acres

Larger value = 179.58 acres