If the normal probability density function \( f(x) \) is given by
\( f(x)=\displaystyle\frac{1}{A\sqrt{2B}}e^{-C\left(\frac{x-D}{2\sigma}\right)^2} \) where \( -\infty\lt x\lt\infty \), the mean and variance of the random variable X are \( \mu \) and \( \sigma^2 \) respectively then
Match List-I with List-II
| List-I | List-II | ||
| (A) | A | (I) | 2 |
| (B) | B | (II) | \( \pi \) |
| (C) | C | (III) | \( \mu \) |
| (D) | D | (IV) | \( \sigma \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
The normal density with mean \( \mu \) and variance \( \sigma^2 \) is
\( \displaystyle f(x)=\frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}} \)
The given exponent can be rewritten as
\( \displaystyle -C\left(\frac{x-D}{2\sigma}\right)^2=-\frac{C(x-D)^2}{4\sigma^2} \)
Comparing the two forms:
(A) A
This matches (IV) \( \sigma \), from \( A\sqrt{2B}=\sigma\sqrt{2\pi} \).
(B) B
This matches (II) \( \pi \), from the same factor.
(C) C
This matches (I) 2:
\( \displaystyle \frac{C}{4\sigma^2}=\frac{1}{2\sigma^2}\ \Rightarrow\ C=2 \)
(D) D
This matches (III) \( \mu \): the exponent is centred at the mean, \( (x-\mu)^2 \).
So the matching is (A)-(IV), (B)-(II), (C)-(I), (D)-(III).