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Question:
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A random variable $\mathrm{X}$ has following distribution
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathrm{X}=\mathrm{x} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{k} & 3 \mathrm{k} & 5 \mathrm{k} & 7 \mathrm{k} & 8 \mathrm{k} & \mathrm{K} \\
\hline
\end{array}
Then $\mathrm{P}(2 \leq \mathrm{x} < 5)=$
Options:
\begin{array}{|l|l|l|l|l|l|l|}
\hline \mathrm{X}=\mathrm{x} & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline \mathrm{P}(\mathrm{X}=\mathrm{x}) & \mathrm{k} & 3 \mathrm{k} & 5 \mathrm{k} & 7 \mathrm{k} & 8 \mathrm{k} & \mathrm{K} \\
\hline
\end{array}
Then $\mathrm{P}(2 \leq \mathrm{x} < 5)=$
Solution:
2806 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{5}$
We have $\mathrm{k}+3 \mathrm{k}+5 \mathrm{k}+7 \mathrm{x}+8 \mathrm{k}+\mathrm{k}=1 \Rightarrow \mathrm{k}=\frac{1}{25}$
$$
\therefore \mathrm{P}(2 \leq \mathrm{x} \leq 5)=\frac{1}{25}(3+5+7)=\frac{15}{25}=\frac{3}{5}
$$
$$
\therefore \mathrm{P}(2 \leq \mathrm{x} \leq 5)=\frac{1}{25}(3+5+7)=\frac{15}{25}=\frac{3}{5}
$$
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