Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | In a binomial distribution, if \( n=16,p=0.75 \), then the mean is | I. | 4 |
| B. | In a binomial distribution, if \( n=25,p=\displaystyle\frac{1}{5} \), then variance is | II. | 32 |
| C. | In a binomial distribution, the mean is 8 and the variance is 6, then the number of trials is | III. | 18 |
| D. | If the mean and variance of a binomial distribution are 6 and 4 respectively, then the number of trials is | IV. | 12 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-II, D-III
For a binomial distribution, mean = np and variance = npq, with q = 1 − p. Dividing, \( q=\frac{\text{variance}}{\text{mean}} \).
(A) n = 16, p = 0.75: the mean
This matches (IV) 12.
\( np=16\times 0.75=12 \)
(B) n = 25, \( p=\frac{1}{5} \): the variance
This matches (I) 4.
\( \displaystyle npq=25\times\frac{1}{5}\times\frac{4}{5}=4 \)
(C) Mean 8, variance 6: the number of trials
This matches (II) 32.
\( \displaystyle q=\frac{6}{8}=\frac{3}{4} \)
\( \displaystyle p=\frac{1}{4} \)
\( \displaystyle n=\frac{8}{1/4}=32 \)
(D) Mean 6, variance 4: the number of trials
This matches (III) 18.
\( \displaystyle q=\frac{4}{6}=\frac{2}{3} \)
\( \displaystyle p=\frac{1}{3} \)
\( \displaystyle n=\frac{6}{1/3}=18 \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).