Match List - I with List - II.
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List - I |
List - II |
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(A) In a binomial distribution, if n = 10, q = 0.25, then its mean is |
(I) 12 |
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(B) If the mean of a binomial distribution is 6 and its variance is 3, then p is |
(II) 7.5 |
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(C) In a binomial distribution, the probability of getting a success is $\frac{1}{4}$ and the standard distribution is 3, then its mean is |
(III) 16 |
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(D) If the mean and variance of a binomial distribution are 4 and 3 respectively, then the number of trials is |
(IV) $\frac{1}{2}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
$\text{(A)}\; n=10,\; q=0.25 \Rightarrow p=0.75$
$\text{Mean} = np = 10 \times 0.75 = 7.5 \Rightarrow \text{matches (II)}$
$\text{(B)}\; \text{Mean}=6,\; \text{Variance}=3$
$np=6,\; npq=3 \Rightarrow q=\frac{3}{6}=\frac{1}{2} \Rightarrow p=\frac{1}{2} \Rightarrow \text{matches (IV)}$
$\text{(C)}\; p=\frac{1}{4},\; \sigma=3 \Rightarrow npq=9$
$n \cdot \frac{1}{4}\cdot \frac{3}{4} = 9 \Rightarrow \frac{3n}{16}=9 \Rightarrow n=48$
$\text{Mean}=np=48 \cdot \frac{1}{4}=12 \Rightarrow \text{matches (I)}$
$\text{(D)}\; \text{Mean}=4,\; \text{Variance}=3$
$np=4,\; npq=3 \Rightarrow q=\frac{3}{4},\; p=\frac{1}{4}$
$n=\frac{4}{1/4}=16 \Rightarrow \text{matches (III)}$
A–II,\; B–IV,\; C–I,\; D–III