X Has the Following Probability Distribution P(X)$$\begin{array} { l l l l l }
\mathrm { X } & 1 & 2 & 3 & 4 \
\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2
\end{array}$$

Question

X Has the Following Probability Distribution P(X)$$\begin{array} { l l l l l } 
\mathrm { X } & 1 & 2 & 3 & 4 \
\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2
\end{array}$$

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The Answer of X Has the Following Probability Distribution P(X)$$\begin{array} { l l l l l } 
\mathrm { X } & 1 & 2 & 3 & 4 \
\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2
\end{array}$$

X Has the Following Probability Distribution P(X) $\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}$

Essay

Correct Answer:
Verified
9125
E[X] = (1)(.1)+ (2)(.5)+ ...
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X Has the Following Probability Distribution P(X) X1234P(X).1.5.2.2\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}XP(X)​1.1​2.5​3.2​4.2​

Essay

Correct Answer:Verified9125E[X] = (1)(.1)+ (2)(.5)+ ...View AnswerUnlock this answer nowGet Access to more Verified Answers free of chargeView Full Answer For Free

X Has the Following Probability Distribution P(X) $\begin{array} { l l l l l } \mathrm { X } & 1 & 2 & 3 & 4 \\\mathrm { P } ( \mathrm { X } ) & .1 & .5 & .2 & .2\end{array}$

Correct Answer:
Verified
9125
E[X] = (1)(.1)+ (2)(.5)+ ...
View Answer
Unlock this answer now
Get Access to more Verified Answers free of charge