Match List-I with List-II
| List-I | List-II | ||
| (A) | For a random variable \( X \), \( E(X)=3 \) and \( E(X^2)=11 \), then \( Var(X) \) is | (I) | 25 |
| (B) | For a random variable \( X \), \( E(X)=3 \) and \( E(X^2)=13 \), then \( Var\left(\displaystyle\frac{X}{2}\right) \) is | (II) | 2 |
| (C) | In a binomial distribution, if n = 20, q = 0.75, then its mean is | (III) | 1 |
| (D) | If in a binomial distribution, mean is 5 and variance is 4, then the number of trials is | (IV) | 5 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
(A) E(X) = 3, E(X2) = 11
This matches (II) 2.
\( \displaystyle Var(X)=E(X^2)-[E(X)]^2 \)
\( \displaystyle =11-9=2 \)
(B) E(X) = 3, E(X2) = 13
This matches (III) 1.
\( \displaystyle Var(X)=13-9=4 \)
\( \displaystyle Var\left(\frac{X}{2}\right)=\frac{1}{4}Var(X)=\frac{4}{4}=1 \)
(C) n = 20, q = 0.75
This matches (IV) 5.
\( \displaystyle p=1-q=0.25 \)
\( \displaystyle \text{Mean}=np=20\times 0.25=5 \)
(D) mean 5, variance 4
This matches (I) 25.
\( \displaystyle np=5 \)
\( \displaystyle npq=4 \)
\( \displaystyle q=\frac{npq}{np}=\frac{4}{5} \)
\( \displaystyle p=\frac{1}{5} \)
\( \displaystyle n=\frac{5}{p}=25 \)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).
Tip: \( Var(aX)=a^2Var(X) \), so halving X divides the variance by 4, not by 2.