If a random variable X has the probability distribution x, 0, 1, 2, 3, 4, 5, 6, 7 | ---|--…

Question

If a random variable $X$ has the probability distribution
 $x$, 0, 1, 2, 3, 4, 5, 6, 7 |
---|---|---|---|---|---|---|---|---|
 $P(x)$, 0, $2k$, $k$, $3k$, $2k^2$, $2k$, $k^2+k$, $7k^2$ |

then $P(3<x\le6)$ is equal to

Accepted Answer

Since $\sum P(x)=1$:

$$2k+k+3k+2k^2+2k+k^2+k+7k^2=1 \implies 10k^2+9k-1=0 \implies (10k-1)(k+1)=0$$

So $k=\dfrac{1}{10}$ (rejecting $k=-1$).

$$P(3<x\le6)=P(4)+P(5)+P(6)=2k^2+2k+(k^2+k)=3k^2+3k = 3(0.01+0.1)=0.33$$

Solutions · 1