A discrete random variable takes integer values from to , and its probability mass function is given bywhere is a nonzero constant. Compute[3 points]\
[THOUGHT] The first step is to determine the value of (a). The sum of all probabilities must equal 1. Therefore: [ \text{P}(X=0) + \text{P}(X=1) + \text{P}(X=2) + \text{P}(X=3) + \text{P}(X=4) = 1 ] Substitute the given probabilities: [ \frac{|2(0) - 1|}{12} + \frac{|2(1) - 1|}{12} + \frac{|2(2) - 1|}{12} + \frac{|2(3) - 1|}{12} + a = 1 ] Calculate the absolute values: [ \frac{1}{12} + \frac{1}{12} + \frac{3}{12} + \frac{5}{12} + a = 1 ] Simplify the fractions: [ \frac{1 + 1 + 3 + 5}{12} + a = 1 ] [ \frac{10}{12} + a = 1 ] [ \frac{5}{6} + a = 1 ] Solve for (a): [ a = 1 - \frac{5}{6} ] [ a = \frac{1}{6} ]
Now that (a) is determined, the next step is to find (\text{E}(X)) and (\text{E}(X^2)).
Calculate (\text{E}(X)): [ \text{E}(X) = \sum_{x=0}^{4} x \cdot \text{P}(X=x) ] Substitute the probabilities: [ \text{E}(X) = 0 \cdot \frac{1}{12} + 1 \cdot \frac{1}{12} + 2 \cdot \frac{3}{12} + 3 \cdot \frac{5}{12} + 4 \cdot \frac{1}{6} ] [ \text{E}(X) = 0 + \frac{1}{12} + \frac{6}{12} + \frac{15}{12} + \frac{8}{12} ] [ \text{E}(X) = \frac{1 + 6 + 15 + 8}{12} ] [ \text{E}(X) = \frac{30}{12} ] [ \text{E}(X) = \frac{5}{2} ]
Calculate (\text{E}(X^2)): [ \text{E}(X^2) = \sum_{x=0}^{4} x^2 \cdot \text{P}(X=x) ] Substitute the probabilities: [ \text{E}(X^2) = 0^2 \cdot \frac{1}{12} + 1^2 \cdot \frac{1}{12} + 2^2 \cdot \frac{3}{12} + 3^2 \cdot \frac{5}{12} + 4^2 \cdot \frac{1}{6} ] [ \text{E}(X^2) = 0 + \frac{1}{12} + \frac{12}{12} + \frac{45}{12} + \frac{16}{6} ] Convert (\frac{16}{6}) to a fraction with denominator 12: [ \text{E}(X^2) = 0 + \frac{1}{12} + \frac{12}{12} + \frac{45}{12} + \frac{32}{12} ] [ \text{E}(X^2) = \frac{1 + 12 + 45 + 32}{12} ] [ \text{E}(X^2) = \frac{90}{12} ] [ \text{E}(X^2) = \frac{15}{2} ]
Now, find the variance of (X): [ \text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2 ] [ \text{Var}(X) = \frac{15}{2} - \left(\frac{5}{2}\right)^2 ] [ \text{Var}(X) = \frac{15}{2} - \frac{25}{4} ] Convert (\frac{15}{2}) to a fraction with denominator 4: [ \text{Var}(X) = \frac{30}{4} - \frac{25}{4} ] [ \text{Var}(X) = \frac{5}{4} ]
Finally, find (\text{V}\left(\frac{1}{a}X\right)): [ \text{V}\left(\frac{1}{a}X\right) = \left(\frac{1}{a}\right)^2 \text{Var}(X) ] Substitute (a = \frac{1}{6}) and (\text{Var}(X) = \frac{5}{4}): [ \text{V}\left(\frac{1}{a}X\right) = \left(\frac{1}{\frac{1}{6}}\right)^2 \cdot \frac{5}{4} ] [ \text{V}\left(\frac{1}{a}X\right) = 6^2 \cdot \frac{5}{4} ] [ \text{V}\left(\frac{1}{a}X\right) = 36 \cdot \frac{5}{4} ] [ \text{V}\left(\frac{1}{a}X\right) = 45 ]
Therefore, the variance is: [ \boxed{45} ] [/THOUGHT]