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Question 6

If a random variable x has the probability distribution

then $$ P(3< x\leq 6)$$ is equal to

Solution

$$P(x)=0,,2k,,k,,3k,,2k^2,,2k,,k^2+k,,7k^2$$

Total probability sum must be 1.

$$10k^2+9k=1$$

Solve:
$$10k^2+9k-1=0$$
$$k=\frac{-9\pm\sqrt{81+40}}{20}=\frac{-9\pm11}{20}$$

k = 0.1
P(3$$<$$ x $$\le6$$) = P(4) + P(5) + P(6)

$$(P(4)=2k^2=2(0.1)^2=0.02)$$
$$(P(5)=2k=0.2)$$
$$(P(6)=k^2+k=0.01+0.1=0.11)$$

Sum: 0.02 + 0.2 + 0.11 = 0.33

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