The probability distribution for X is as follows: $$\begin{array} { l | r r r r }
x & - 1 & 0 & 1 & 2 \
\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1
\end{array}$$ a. Find E[5X + 1].
b. Find V[5X + 1].

Question

The probability distribution for X is as follows: $$\begin{array} { l | r r r r } 
x & - 1 & 0 & 1 & 2 \
\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1
\end{array}$$ a. Find E[5X + 1].
b. Find V[5X + 1].

Quizplus · Accepted Answer

The Answer of The probability distribution for X is as follows: $$\begin{array} { l | r r r r } 
x & - 1 & 0 & 1 & 2 \
\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1
\end{array}$$ a. Find E[5X + 1].
b. Find V[5X + 1].

The Probability Distribution for X Is as Follows x−1012p(x)0.10.250.550.1\begin{array} { l | r r r r } x & - 1 & 0 & 1 & 2 \\\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1\end{array}xp(x)​−10.1​00.25​10.55​20.1​​

The Probability Distribution for X Is as Follows $\begin{array} { l | r r r r } x & - 1 & 0 & 1 & 2 \\\hline p ( x ) & 0.1 & 0.25 & 0.55 & 0.1\end{array}$