If a random variable $x$ has the probability distribution x 0 1 2 3 4 5 6 7 $$p(x)$$ 0 2 k k 3 k $$2

Question

If a random variable $x$ has the probability distribution x 0 1 2 3 4 5 6 7 $$p(x)$$ 0 2 k k 3 k $$2k^{2}$$ 2 k $$k^{2} + k$$ $$7k^{2}$$ then $P(3 < x \leq 6)$ is equal to

Accepted Answer

$\ \sum P\left( x_{i} \right) = 1$$${\Rightarrow 9k + 10k^{2} = 1 }{\Rightarrow 10k^{2} + 9k - 1 = 0 \Rightarrow k = \frac{1}{10} }{P(3 < x \leq 6) = 3k + 3k^{2} }{= \frac{3}{10} + \frac{3}{100} = 0.33 }{= 0.33 }$$

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