The probability distribution for X ,daily demand of a particular newspaper at a local newsagency,( in hundreds) is as follows: $$\begin{array} { l | r r r r }
x & 1 & 2 & 3 & 4 \
\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09
\end{array}$$ a. Find and interpret the expected value of X.
b. Find V(X).
c. Find $\sigma$.

Question

The probability distribution for X ,daily demand of a particular newspaper at a local newsagency,( in hundreds) is as follows: $$\begin{array} { l | r r r r } 
x & 1 & 2 & 3 & 4 \
\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09
\end{array}$$ a. Find and interpret the expected value of X.
b. Find V(X).
c. Find $\sigma$.

Quizplus · Accepted Answer

The Answer of The probability distribution for X ,daily demand of a particular newspaper at a local newsagency,( in hundreds) is as follows: $$\begin{array} { l | r r r r } 
x & 1 & 2 & 3 & 4 \
\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09
\end{array}$$ a. Find and interpret the expected value of X.
b. Find V(X).
c. Find $\sigma$.

The Probability Distribution for X ,Daily Demand of a Particular x1234p(x)0.050.420.440.09\begin{array} { l | r r r r } x & 1 & 2 & 3 & 4 \\\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09\end{array}xp(x)​10.05​20.42​30.44​40.09​​

The Probability Distribution for X ,Daily Demand of a Particular $\begin{array} { l | r r r r } x & 1 & 2 & 3 & 4 \\\hline p ( x ) & 0.05 & 0.42 & 0.44 & 0.09\end{array}$