Answer to Question #203075 in Statistics and Probability for Erick

Question

A random variable X has the following probability distribution: x 0 1 3 4 5 6 7 P X x ( )  0 k 2k 3k k^2 2k^2 7k^2+ k  A. Find k. 2 B. Evaluate P (1<=X< 6).

Answer 1

We can find $k$ from equality: $1=\sum_{k=0}^7P(X=k)=k+2k+3k+k^2+2k^2+7k^2+k=7k+10k^2$ . We solve: $10k^2+7k-1=0$. $k=\frac{-7\pm\sqrt{89}}{20}.$ $k_1\approx-0.822,$ $k_2\approx0.122$.

$P(X\leq1<6)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=k+2k+3k+k^2+2k^2=6k+3k^2$