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Probability Distribution

X 1 2 3 4 5 6...

Question

X 1 2 3 4 5 6
P(X=x) k 2k 3k 4k 5k 6k
The probability distribution of discrete r.v. $X$ is
Then $P (X \leq 4)$ =

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Solution

Sum of the probabilities is equal to 1

$∴ k + 2 k + 3 k + 4 k + 5 k + 6 k = 1$

$21 k = 1$

$k = \frac{1}{21}$

$P (X \leq 4)$

$= P (1) + P (2) + P (3) + P (4)$

$= k + 2 k + 3 k + 4 k$

$= \frac{1}{21} + \frac{2}{21} + \frac{3}{21} + \frac{4}{21}$

$= \frac{1 + 2 + 3 + 4}{21}$

$= \frac{10}{21}$

$∴ P (X \leq 4) = 0.476$

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