Match List-I with List-II
| List-I | List-II | ||
| (A). | The mean of the Binomial distribution \( \mathrm{B}\left(4,\displaystyle\frac{2}{3}\right) \) is | (I). | 4 |
| (B). | The variance of the Binomial distribution \( \mathrm{B}\left(16,\displaystyle\frac{1}{2}\right) \) is | (II). | 2 |
| (C). | The standard deviation of the Binomial distribution \( \mathrm{B}\left(16,\displaystyle\frac{1}{2}\right) \) is | (III). | 9 |
| (D). | The mean of the number of heads in 18 tosses of a coin is | (IV). | \( \displaystyle\frac{8}{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
For a binomial distribution \( \mathrm{B}(n,p) \): mean \( =np \), variance \( =npq \) and standard deviation \( =\sqrt{npq} \).
(A) The mean of \( \mathrm{B}\left(4,\frac{2}{3}\right) \)
This matches (IV) \( \frac{8}{3} \).
The mean is \( 4\times\frac{2}{3}=\frac{8}{3} \).
(B) The variance of \( \mathrm{B}\left(16,\frac{1}{2}\right) \)
This matches (I) 4.
The variance is \( 16\times\frac{1}{2}\times\frac{1}{2}=4 \).
(C) The standard deviation of \( \mathrm{B}\left(16,\frac{1}{2}\right) \)
This matches (II) 2.
The standard deviation is \( \sqrt{4}=2 \).
(D) The mean of the number of heads in 18 tosses of a coin
This matches (III) 9.
The number of heads in 18 tosses is \( \mathrm{B}\left(18,\frac{1}{2}\right) \), with mean \( 18\times\frac{1}{2}=9 \).
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).