[THOUGHT] We need to compute V( (1/a) X ), where V denotes variance. Since variance scales by the square of the constant factor: V(cX) = c^2 V(X). So V( (1/a) X ) = (1/a)^2 V(X) = V(X) / a^2.

First we need to find the constant a. The PMF must sum to 1 over x = 0,1,2,3,4. So sum_{x=0}^3 |2x - 1| / 12 + a = 1.

Compute sum_{x=0}^3 |2x - 1| / 12.

x=0: |20 - 1| = | -1 | = 1 → probability 1/12. x=1: |21 - 1| = |1| = 1 → probability 1/12. x=2: |22 - 1| = |3| = 3 → probability 3/12. x=3: |23 - 1| = |5| = 5 → probability 5/12.

Sum = (1+1+3+5)/12 = 10/12 = 5/6.

So total probability from x=0..3 is 5/6. Then a = 1 - 5/6 = 1/6.

Now we have PMF: P(X=0) = 1/12 P(X=1) = 1/12 P(X=2) = 3/12 = 1/4 P(X=3) = 5/12 P(X=4) = 1/6 = 2/12? Actually 1/6 = 2/12. Let's keep fractions consistent with denominator 12: 1/12, 1/12, 3/12, 5/12, 2/12. Sum = (1+1+3+5+2)/12 = 12/12 = 1. Good.

Now compute E[X] and E[X^2] to get variance V(X) = E[X^2] - (E[X])^2.

Then compute V(X) / a^2 = V(X) / (1/6)^2 = V(X) / (1/36) = 36 * V(X).

We'll compute numerically using Python to avoid mistakes. [/THOUGHT] [PYTHON] import fractions from fractions import Fraction

Probabilities as Fractions

p0 = Fraction(1, 12) p1 = Fraction(1, 12) p2 = Fraction(3, 12) p3 = Fraction(5, 12) p4 = Fraction(1, 6)

Check sum

psum = p0 + p1 + p2 + p3 + p4 print("Sum of probabilities:", psum)

Expected value E[X]

E_X = 0p0 + 1p1 + 2p2 + 3p3 + 4*p4 print("E[X] =", E_X, "=", float(E_X))

Expected value E[X^2]

E_X2 = (02)*p0 + (12)*p1 + (22)*p2 + (32)*p3 + (4**2)*p4 print("E[X^2] =", E_X2, "=", float(E_X2))

Variance V(X)

V_X = E_X2 - E_X**2 print("V(X) =", V_X, "=", float(V_X))

a = 1/6

a = Fraction(1, 6)

Variance of (1/a) X = V(X) / a^2

result = V_X / (a**2) print("V((1/a)X) =", result, "=", float(result)) print("Result as integer? (if fraction denominator 1):", result) [/PYTHON]

[THOUGHT] The Python computation shows that the variance of (1/a)X is exactly 45. This is an integer, so the final answer is 45. [/THOUGHT] \boxed{45}