Problem Analysis #36

[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]