Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | In a binomial distribution, if n = 20, q = 0.75 then its mean is | I. | 25 |
| B. | An unbiased coin is tossed 4 times. The mean of the number of tails is | II. | 5 |
| C. | In a binomial distribution, mean is 5 and variance is 4, then the number of trials is | III. | 0.5 |
| D. | In a binomial distribution, if mean is 8 and variance is 4, then the probability of success is | IV. | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-IV, C-I, D-III
For a binomial distribution, mean = np and variance = npq, with p + q = 1.
(A) n = 20, q = 0.75
This matches (II) 5.
\( p=1-0.75=0.25 \)
\( \text{mean}=np=20\times 0.25=5 \)
(B) An unbiased coin tossed 4 times
This matches (IV) 2.
\( \displaystyle n=4,\ p=\frac{1}{2} \)
\( \displaystyle \text{mean}=4\times\frac{1}{2}=2 \)
(C) mean 5, variance 4
This matches (I) 25.
\( \displaystyle q=\frac{npq}{np}=\frac{4}{5}=0.8 \)
\( p=1-0.8=0.2 \)
\( \displaystyle n=\frac{5}{0.2}=25 \)
(D) mean 8, variance 4
This matches (III) 0.5.
\( \displaystyle q=\frac{npq}{np}=\frac{4}{8}=0.5 \)
\( p=1-0.5=0.5 \)
So the matching is A-II, B-IV, C-I, D-III.