Consider the Population Described by the Probability Distribution Below$\begin{array}{c|c|c|c}
x & 0 & 2 & 4 \
\hline p(x) & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}
\end{array}$

Question

Examlex · Accepted Answer

The Answer is a.

\quad \mu = E ( x ) = \frac { 1 } { 3 } ( 0 ) + \frac { 1 } { 3 } ( 2 ) + \frac { 1 } { 3 } ( 4 ) = 2

b.

\begin{array}{c|c|c|c|c|c|c|c}\bar{x} & 0 & \frac{2}{3} & \frac{4}{3} & 2 & \frac{8}{3} & \frac{10}{3} & 4 \\hline p(\bar{x}) & \frac{1}{27} & \frac{1}{9} & \frac{2}{9} & \frac{7}{27} & \frac{2}{9} & \frac{1}{9} & \frac{1}{27}\end{array}

c.

\begin{array}{c|c|c|c}M & 0 & 2 & 4 \\hline p(M) & \frac{7}{27} & \frac{13}{27} & \frac{7}{27}\end{array}

d.

E ( \bar { x } ) = \frac { 1 } { 27 } ( 0 ) + \frac { 1 } { 9 } \left( \frac { 2 } { 3 } ight) + \frac { 2 } { 9 } \left( \frac { 4 } { 3 } ight) + \frac { 7 } { 27 } ( 2 ) + \frac { 2 } { 9 } \left( \frac { 8 } { 3 } ight) + \frac { 1 } { 9 } \left( \frac { 10 } { 3 } ight) + \frac { 1 } { 27 } ( 4 ) = 2

; Since

E ( \bar { x } ) = \mu

, the sample
mean is an unbiased estimator of

\mu

.

E ( M ) = \frac { 7 } { 27 } ( 0 ) + \frac { 13 } { 27 } ( 2 ) + \frac { 7 } { 27 } ( 4 ) = 2

; Since

E ( M ) = \mu

, the sample median is an unbiased estimator
of

\mu

.
e.

\quad \sigma \frac { 2 } { x } = \frac { 1 } { 27 } ( 0 - 2 ) ^ { 2 } + \frac { 1 } { 9 } \left( \frac { 2 } { 3 } - 2 ight) ^ { 2 } + \frac { 2 } { 9 } \left( \frac { 4 } { 3 } - 2 ight) ^ { 2 } + \frac { 7 } { 27 } ( 2 - 2 ) ^ { 2 } + \frac { 2 } { 9 } \left( \frac { 8 } { 3 } - 2 ight) ^ { 2 }

+ \frac { 1 } { 9 } \left( \frac { 10 } { 3 } - 2 ight) ^ { 2 } + \frac { 1 } { 27 } ( 4 - 2 ) ^ { 2 } = \frac { 8 } { 9 }

\sigma _ { M } ^ { 2 } = \frac { 7 } { 27 } ( 0 - 2 ) ^ { 2 } + \frac { 13 } { 27 } ( 2 - 2 ) ^ { 2 } + \frac { 7 } { 27 } ( 4 - 2 ) ^ { 2 } = \frac { 56 } { 27 }

f. sample mean; The variance is smaller.

Consider the Population Described by the Probability Distribution Below $\begin{array}{c|c|c|c}x & 0 & 2 & 4 \\\hline p(x) & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{array}$

Definitions:

Consider the Population Described by the Probability Distribution Below x024p(x)131313\begin{array}{c|c|c|c}x & 0 & 2 & 4 \\\hline p(x) & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{array}xp(x)​031​​231​​431​​​

Definitions:

Consider the Population Described by the Probability Distribution Below $\begin{array}{c|c|c|c}x & 0 & 2 & 4 \\\hline p(x) & \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{array}$