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  • For the following probability distribution: X 4 E(X) is equal to: A. 0 B. –1 C. –2 D. –1.8

For the following probability distribution:
\begin{array}{|l|l|l|l|l|l|} \hline \mathrm{X} & -4 & -3 & -2 & -1 & 0 \\ \hline \mathrm{P}(\mathrm{X}) & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \\ \hline \end{array}
E(X) is equal to:
A. 0
B. –1
C. –2
D. –1.8

Answers (1)

Given-

 Probability distribution table

\\ \mathrm{E}(\mathrm{X})=\sum \mathrm{X.P}(\mathrm{X}) \\ \mathrm{E}(\mathrm{X})=[(-4) \times(0.1)+(-3) \times(0.2)+(-2) \times(0.3)+(-1) \times(0.2)+(0 \times 0.2)] \\ \mathrm{E}(\mathrm{X})=[-0.4-0.6-0.6-0.2+0] \\ \mathrm{E}(\mathrm{X})=[-1.8] \\ \text { Hence, } \mathrm{E}(\mathrm{X})=-1.8

Option D is correct.

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infoexpert22

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