# Central Limit Theorem: CLT for Averages

#### Lucyzzz

In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 52 acres. Suppose we randomly survey 45 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.)

 acres (smaller value) acres (larger value)

If this is for a normal distribution and you have access to a table for z-scores, consider that if you find the z-score whose area to the left is 25% and the other z-score whose area to the left is 75%, these z-scores represent the upper and lower bounds for the "middle 50%" as desired.

You should then be able to use these z-scores to determine the small and larger value x. z = (x – μ) / (σ / √n)

#### joshuaa

Q1 = 138.9266 = 138.93
Q2 = 174
Q3 = 209.0734 = 209.07

IQR lies between 138.93 and 209.07

#### Lucyzzz

Hello I tried that way and used 138.93 and 209.07, but the computer say is wrong. could please explain more

#### joshuaa

Q1 = 174 - 0.67448x52 = 138.92
Q3 = 174 + 0.67448x52 = 209.07

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Suppose we randomly survey 45 farmers from 1940.

I think that because of this sentence the problem needs an extra attention

Let us try to use the formula MacstersUndead mentioned

z = (x – μ) / (σ / √n)

then

Q1 = 174 - (0.67448x52)/√45 = 168.77
Q3 = 174 + (0.67448x52)/√45 = 179.23

What the computer says now?

Lucyzzz

#### Lucyzzz

Q1 = 174 - 0.67448x52 = 138.92
Q3 = 174 + 0.67448x52 = 209.07

---------------------------------------------------------------------------------------------
Suppose we randomly survey 45 farmers from 1940.

I think that because of this sentence the problem needs an extra attention

Let us try to use the formula MacstersUndead mentioned

z = (x – μ) / (σ / √n)

then

Q1 = 174 - (0.67448x52)/√45 = 168.77
Q3 = 174 + (0.67448x52)/√45 = 179.23

What the computer says now?

#### Lucyzzz

You are a Genious, Joshuaa. Thank you so much

joshuaa