12. Let X be a normal random variable with mean 678 and standard deviation 8.85. Let Q₁ the probability that X2 656, Q2 the probability that X < 685, and Q3 the probability that 664 < X < 679. Let Q = ln(3+1Q1 +21Q21 +3|Q3). Then T = 5 sin² (1000) satisfies:- (A) 0 ≤ T < 1. (E) 4≤ T ≤ 5. (B) 1 ≤ T < 2. (C) 2 ≤ T < 3. (D) 3 ≤ T < 4. - -

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