Match List-I with List-II
| List-I | List-II | ||
| (A) | Mean of a Poisson variate X for which \(P(X = 2) = P(X = 3)\) | (I) | 0.5 |
| (B) | Standard normal variate corresponding to X = 62, when X is normal with mean 50 and standard deviation 8 | (II) | 3 |
| (C) | Standard deviation of a Poisson variate with mean 16 | (III) | 1.5 |
| (D) | \(P(X \gt \mu)\) for a normal variate X with mean \(\mu\) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A) Mean of a Poisson variate X for which \(P(X = 2) = P(X = 3)\)
This matches (II) 3. With mean m (m > 0), using the Poisson probability \(P(X = r) = \frac{e^{-m}\,m^{r}}{r!}\):
\( \displaystyle \frac{e^{-m}\,m^{2}}{2!} = \frac{e^{-m}\,m^{3}}{3!} \)
\( \displaystyle \frac{1}{2} = \frac{m}{6} \)
\(m = 3\)
(B) Standard normal variate corresponding to X = 62
This matches (III) 1.5.
\( \displaystyle Z = \frac{62 - 50}{8} = \frac{12}{8} = 1.5 \)
(C) Standard deviation of a Poisson variate with mean 16
This matches (IV) 4. The variance equals the mean, 16, so
\(\sigma = \sqrt{16} = 4\)
(D) \(P(X \gt \mu)\) for a normal variate
This matches (I) 0.5. The normal curve is symmetric about the mean, so half of the total area 1 lies to the right of \(\mu\).