Match List-I with List-II
| List-I | List-II | ||
| (A) | Mean of a normal variate X is 12 and standard deviation is 4, then the Z-score of data point 20 is | (I) | 1 |
| (B) | If X is Poisson variate such that P(X=1) = 2P(X=2), then the mean of X is | (II) | 8 |
| (C) | In a binomial distribution, if mean is 5 and variance is 4, then the number of trials is | (III) | 2 |
| (D) | In a binomial distribution, if n = 20 and q = 0.6, then its mean is | (IV) | 25 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
(A) Mean of a normal variate X is 12 and standard deviation is 4, then the Z-score of data point 20 is
This matches (III) 2.
\( \displaystyle Z=\frac{X-\mu}{\sigma}=\frac{20-12}{4}=2 \)
(B) If X is Poisson variate such that P(X=1) = 2P(X=2), then the mean of X is
This matches (I) 1.
\( \displaystyle \frac{e^{-\lambda}\lambda}{1!}=2\cdot\frac{e^{-\lambda}\lambda^2}{2!} \)
\( \displaystyle \lambda=\lambda^2 \)
\( \displaystyle \lambda=1\quad(\lambda\gt 0) \)
The mean of a Poisson variate is \( \lambda=1 \).
(C) In a binomial distribution, if mean is 5 and variance is 4, then the number of trials is
This matches (IV) 25.
\( \displaystyle q=\frac{npq}{np}=\frac{4}{5} \)
\( \displaystyle p=1-\frac{4}{5}=\frac{1}{5} \)
\( \displaystyle n=\frac{np}{p}=\frac{5}{1/5}=25 \)
(D) In a binomial distribution, if n = 20 and q = 0.6, then its mean is
This matches (II) 8.
\( \displaystyle p=1-0.6=0.4 \)
\( \displaystyle np=20\times 0.4=8 \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).