Question

A random variable X has the following probability mass function X 2 3 4 5 6 7 8 P(X=x) 1/16 2k 3/16 1/4 3/16 1/8 k i) Find k, hence determine the mean of the distribution. ii) Obtain �3≤�5) iii) Determine F(5)

Answer 1

i)

$$\dfrac{1}{16}+2k+\dfrac{3}{16}+\dfrac{1}{4}+\dfrac{3}{16}+\dfrac{1}{8}+k=1$$

$$3k=\dfrac{3}{16}$$

$$k=\dfrac{1}{16}$$

$$E(X)=2(\dfrac{1}{16})+ 3(\dfrac{2}{16})+ 4(\dfrac{3}{16})+ 5(\dfrac{4}{16})$$

$$+ 6(\dfrac{3}{16})+ 7(\dfrac{2}{16})+ 8(\dfrac{1}{16})=5$$

ii)

$$P(3\leq X\leq 5)=P(X=3)+P(X=4)$$

$$+P(X=5)=\dfrac{2}{16}+ \dfrac{3}{16}+ \dfrac{4}{16}=\dfrac{9}{16}$$

iii)

$$F(5)=f(2)+f(3)+f(4)+f(5)$$

$$=\dfrac{1}{16}+\dfrac{2}{16}+ \dfrac{3}{16}+ \dfrac{4}{16}=\dfrac{5}{8}$$