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] To compute the variance of (\dfrac{1}{a}X), we first determine the constant (a) by ensuring the probability mass function sums to 1. Then, we calculate the expected value (E[X]) and (E[X^2]) to find (\text{Var}(X)). Finally, using the property (\text{Var}(cX) = c^2 \text{Var}(X)), we compute the desired variance.
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Find (a):
- Sum probabilities for (x = 0, 1, 2, 3): [ \sum_{x=0}^3 \frac{|2x - 1|}{12} = \frac{1}{12} + \frac{1}{12} + \frac{3}{12} + \frac{5}{12} = \frac{10}{12} = \frac{5}{6} ]
- Solve ( \frac{5}{6} + a = 1 ) to get ( a = \frac{1}{6} ).
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Calculate (E[X]): [ E[X] = \sum_{x=0}^4 x \cdot P(X = 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} = \frac{5}{2} ]
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Calculate (E[X^2]): [ E[X^2] = \sum_{x=0}^4 x^2 \cdot P(X = x) = 0 \cdot \frac{1}{12} + 1 \cdot \frac{1}{12} + 4 \cdot \frac{3}{12} + 9 \cdot \frac{5}{12} + 16 \cdot \frac{1}{6} = \frac{15}{2} ]
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Compute (\text{Var}(X)): [ \text{Var}(X) = E[X^2] - (E[X])^2 = \frac{15}{2} - \left(\frac{5}{2}\right)^2 = \frac{5}{4} ]
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Compute (\text{Var}\left(\dfrac{1}{a}X\right)): [ \text{Var}\left(\dfrac{1}{a}X\right) = \left(\dfrac{1}{a}\right)^2 \text{Var}(X) = 6^2 \cdot \frac{5}{4} = 36 \cdot \frac{5}{4} = 45 ]
[/THOUGHT] [PYTHON]
Calculate the constant a
prob_sum = sum(abs(2*x - 1)/12 for x in range(4)) a = 1 - prob_sum
Calculate E[X]
E_X = sum(x * (abs(2*x - 1)/12 if x < 4 else a) for x in range(5))
Calculate E[X^2]
E_X2 = sum(x**2 * (abs(2*x - 1)/12 if x < 4 else a) for x in range(5))
Variance of X
var_X = E_X2 - E_X**2
Variance of (1/a)X
var_scaled = (1/a)**2 * var_X
print(var_scaled) [/PYTHON]
[THOUGHT] The Python output is approximately 45, which matches our manual calculation. The variance of (\dfrac{1}{a}X) is (\boxed{45}). [/THOUGHT]