Question

In 1940 the average size of a U.S. farm was 174 acres†. Let's say that the standard deviation was 56 acres. Suppose we randomly survey 46 farmers from 1940.

• Part (c)

  Give the distribution of

  X.

  (Round your standard deviation to two decimal places.)

  X

  ~ ,

• Part (d)

  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.)

  acres	(smaller value)
  acres	(larger value)

Answer 1

Solution:-

Given that,

mean = = 174

standard deviation = = 56

n = 46

= = 174

= / n = 56 / 46 = 8.26

The sampling distribution of is approximately normal   N ( 174, 8.26 )

Using standard normal table,

P( -z < Z < z) = 50%

= P(Z < z) - P(Z <-z ) = 0.50

= 2P(Z < z) - 1 = 0.50

= 2P(Z < z) = 1 + 0.550

= P(Z < z) = 1.50 / 2

= P(Z < z) = 0.75

= P(Z < 0.6745) = 0.75

= z  ± 0.6745

Using z-score formula  

= z * +   

= -0.6745 * 8.26 + 174

= 168.43

Using z-score formula  

= z * +   

= 0.6745 * 8.26 + 174

= 179.57

smaller value = 168.43

larger value = 179.57