Match List-I with List-II
| List-I | List-II |
| (A) The mean of the Binomial distribution \( B\left(10,\displaystyle\frac{1}{5}\right) \) is | (I) 1.6 |
| (B) The Variance of the Binomial distribution \( B\left(12,\displaystyle\frac{1}{2}\right) \) is | (II) 10 |
| (C) The standard deviation of the Binomial distribution \( B\left(16,\displaystyle\frac{1}{5}\right) \) is | (III) 2 |
| (D) The mean of the Binomial distribution \( B\left(25,\displaystyle\frac{2}{5}\right) \) is | (IV) 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
For B(n, p): mean = np, variance = npq and standard deviation = \( \sqrt{npq} \), where q = 1 − p.
(A) The mean of \( B\left(10,\frac{1}{5}\right) \)
This matches (III) 2.
\( \displaystyle 10\times\frac{1}{5}=2 \)
(B) The variance of \( B\left(12,\frac{1}{2}\right) \)
This matches (IV) 3.
\( \displaystyle 12\times\frac{1}{2}\times\frac{1}{2}=3 \)
(C) The standard deviation of \( B\left(16,\frac{1}{5}\right) \)
This matches (I) 1.6.
\( \displaystyle \sqrt{16\times\frac{1}{5}\times\frac{4}{5}}=\sqrt{\frac{64}{25}} \)
\( \displaystyle =\frac{8}{5}=1.6 \)
(D) The mean of \( B\left(25,\frac{2}{5}\right) \)
This matches (II) 10.
\( \displaystyle 25\times\frac{2}{5}=10 \)
So the matching is (A) - (III), (B) - (IV), (C) - (I), (D) - (II).