MHT CET · Maths · Statistics

For the probability distribution

Then the \(\operatorname{Var}(\mathrm{X})\) is
(Given :
\(\left.(0.25)^2=0.0625,(0.35)^2=0 \cdot 1225,(0.45)^2=0 \cdot 2025\right)\)

A 0.8275
B 1.1225
C 1.8275
D \(2 \cdot 0725\)

Verified Solution

Answer & Solution

Correct Answer

(C) 1.8275

Step-by-step Solution

Detailed explanation

\(E(X)= (-2)(0.1)+(-1)(0.2)+0(0.2)+(1)(0.3) \) \( +2(0.15)+3(0.05) \)
\( = -0.2-0.2+0+0.3+0.3+0.15 \)
\( = 0.35\)
\(\therefore \operatorname{Var}(\mathrm{X}) =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \)
\( =(-2)^2(0.1)+(-1)^2(0.2)+0^2(0.2) \) \( +1^2(0.3)+2^2(0.15)+3^2(0.05)-(0.35)^2 \)
\( =0.4+0.2+0+0.3+0.6\)
\(=1.95-(0.35)^2 \)
\( =1.95-0.1225 \)
\( =1.8275\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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For the probability distributionThen the \(\operatorname{Var}(\mathrm{X})\) is(Given :\(\left.(0.25)^2=0.0625,(0.35)^2=0 \cdot 1225,(0.45)^2=0 \cdot 2025\right)\)

Step-by-step Solution

For the probability distribution

Then the \(\operatorname{Var}(\mathrm{X})\) is
(Given :
\(\left.(0.25)^2=0.0625,(0.35)^2=0 \cdot 1225,(0.45)^2=0 \cdot 2025\right)\)