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.

Give the distribution of
XBar

xbar ~ N ( , )

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.

Give the distribution of 
XBar 

xbar ~ N (      ,       )

anonymous · Answer

$$\bar{X}=\frac{1}{n}\sum_{i=1}^nX_i$$
$\bar{X}$ will be normally distributed with mean $E(\bar{X})$ and variance $V(\bar{X})$. Suppose $E(X_i)=\mu$ and $V(X_i)=\sigma^2$, then
$$\begin{align*}E(\bar{X})&=E\left(\frac{1}{n}\sum_{i=1}^nX_i\right)\
&=\frac{1}{n}E\left(\sum_{i=1}^nX_i\right)\
&=\frac{1}{n}\sum_{i=1}^nE\left(X_i\right)\
&=\frac{1}{n}\sum_{i=1}^n\mu\
&=\frac{1}{n}(n\mu)\\
&=\mu\end{align*}$$
$$\begin{align*}V(\bar{X})&=V\left(\frac{1}{n}\sum_{i=1}^nX_i\right)\
&=\frac{1}{n^2}\sum_{i=1}^nV\left(X_i\right)\
&=\frac{1}{n^2}(n\sigma^2)\
&=\frac{\sigma^2}{n}\
SD(\bar{X})&=\sqrt{\frac{\sigma^2}{n}}\
&=\frac{\sigma}{\sqrt n}
\end{align*}$$
You're given that $\mu=174$, $\sigma=55$ and $n=38$, so
$$\bar{X}\sim N\left(\mu,~\frac{\sigma}{\sqrt n}\right)$$