The probability function of a discrete random variable X is as follows: Values of X: x -4 -2 0 2 4 P(x) k 2k 3k 5k 6k i) Find the value of k. ii) Find the probability of the value of X exactly one. iii) Find the probability of the value of X between -2 and 2. iv) Estimate expected value and standard deviation of X.

i)

$\sum_ip_i=1$

$k +2k+ 3k+ 5k+ 6k=1$

$k=1/17$

ii)

$P(X=1)=0$

iii)

$P(-2\leq X\leq2)=P(X=-2)+P(X=0)$

$+P(X=2)=2/17+3/17+5/17=10/17$

$P(-2<X<2)=P(X=0)=3/17$

iv)

$E(X)=\dfrac{1}{17}(-4)+\dfrac{2}{17}(-2)+\dfrac{3}{17}(0)+\dfrac{5}{17}(2)$

$+\dfrac{6}{17}(4)=\dfrac{26}{17}$

$E(X^2)=\dfrac{1}{17}(-4)^2+\dfrac{2}{17}(-2)^2+\dfrac{3}{17}(0)^2$

$+\dfrac{5}{17}(2)^2+\dfrac{6}{17}(4)^2=\dfrac{140}{17}$

$Var(X)=\sigma^2=E(X^2)-(E(X))^2$

$=\dfrac{140}{17}-(\dfrac{26}{17})^2=\dfrac{1704}{289}$

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{1704}{289}}=\dfrac{2\sqrt{71}}{17}$