Question

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

Solution

## Sum of the probabilities is equal to 1$$\therefore k+2k+3k+4k+5k+6k=1$$$$21k=1$$$$k=\dfrac{1}{21}$$$$P(X\leq 4)$$$$=P(1)+P(2)+P(3)+P(4)$$$$=k+2k+3k+4k$$$$=\dfrac{1}{21}+\dfrac{2}{21}+\dfrac{3}{21}+\dfrac{4}{21}$$$$=\dfrac{1+2+3+4}{21}$$$$=\dfrac{10}{21}$$$$\therefore P(X\leq 4)=0.476$$Applied Mathematics

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