If \( X \) is a normal variate with mean 12 and standard deviation 4 and \( Z \) is standard normal variate then match the following
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( P(X\ge 20) \) | (I) | \( 0.5-P(0\le Z\le 3) \) |
| (B) | \( P(0\le X\le 12) \) | (II) | \( 0.5+P(0\le Z\le 2) \) |
| (C) | \( P(X\le 0) \) | (III) | \( P(0\le Z\le 3) \) |
| (D) | \( P(X\le 20) \) | (IV) | \( 0.5-P(0\le Z\le 2) \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)
Standardising with \( Z=\frac{X-12}{4} \): X = 20 gives Z = 2, X = 12 gives Z = 0 and X = 0 gives Z = −3. The standard normal curve is symmetric about 0, with area 0.5 on each side.
(A) \( P(X\ge 20) \)
This matches (IV) \( 0.5-P(0\le Z\le 2) \).
\( \displaystyle P(Z\ge 2)=0.5-P(0\le Z\le 2) \)
(B) \( P(0\le X\le 12) \)
This matches (III) \( P(0\le Z\le 3) \).
\( \displaystyle P(-3\le Z\le 0)=P(0\le Z\le 3)\quad\text{(symmetry)} \)
(C) \( P(X\le 0) \)
This matches (I) \( 0.5-P(0\le Z\le 3) \).
\( \displaystyle P(Z\le -3)=P(Z\ge 3) \)
\( \displaystyle =0.5-P(0\le Z\le 3) \)
(D) \( P(X\le 20) \)
This matches (II) \( 0.5+P(0\le Z\le 2) \).
\( \displaystyle P(Z\le 2)=P(Z\le 0)+P(0\le Z\le 2) \)
\( \displaystyle =0.5+P(0\le Z\le 2) \)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).