The pmf for X = the number of major defects on a randomly selected gas stove of a certain type is $$\begin{array} { l c c c c c }
\hline x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & .10 & .15 & .45 & .25 & .05 \
\hline
\end{array}$$ Compute the following:
a. E(X)
b. V(X) directly from the definition
c The standard deviation of X
d. V(X) using the shortcut formula

Question

The pmf for X = the number of major defects on a randomly selected gas stove of a certain type is $$\begin{array} { l c c c c c } 
\hline x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & .10 & .15 & .45 & .25 & .05 \
\hline
\end{array}$$ Compute the following: 
a. E(X)
b. V(X) directly from the definition
c The standard deviation of X
d. V(X) using the shortcut formula

Quizplus · Accepted Answer

The Answer of The pmf for X = the number of major defects on a randomly selected gas stove of a certain type is $$\begin{array} { l c c c c c } 
\hline x & 0 & 1 & 2 & 3 & 4 \
\hline P ( x ) & .10 & .15 & .45 & .25 & .05 \
\hline
\end{array}$$ Compute the following: 
a. E(X)
b. V(X) directly from the definition
c The standard deviation of X
d. V(X) using the shortcut formula

The Pmf for X = the Number of Major Defects x01234P(x).10.15.45.25.05\begin{array} { l c c c c c } \hline x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & .10 & .15 & .45 & .25 & .05 \\\hline\end{array}xP(x)​0.10​1.15​2.45​3.25​4.05​​

The Pmf for X = the Number of Major Defects $\begin{array} { l c c c c c } \hline x & 0 & 1 & 2 & 3 & 4 \\\hline P ( x ) & .10 & .15 & .45 & .25 & .05 \\\hline\end{array}$