X is a normal variate with mean 200 and standard deviation 20, and Z is the standard normal variate. (Given \(P(0 \le Z \le 0.5) = 0.1915\), \(P(0 \le Z \le 1) = 0.3413\) and \(P(0 \le Z \le 1.5) = 0.4332\))
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(P(X \le 190)\) | (I) | 0.0668 |
| (B) | \(P(190 \le X \le 230)\) | (II) | 0.1498 |
| (C) | \(P(X \ge 230)\) | (III) | 0.3085 |
| (D) | \(P(210 \le X \le 220)\) | (IV) | 0.6247 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Using \(Z = \frac{X - 200}{20}\): X = 190 gives Z = −0.5, X = 210 gives Z = 0.5, X = 220 gives Z = 1 and X = 230 gives Z = 1.5. The curve is symmetric about the mean, with area 0.5 on each side.
(A) \(P(X \le 190)\)
This matches (III) 0.3085.
\(P(Z \le -0.5) = 0.5 - 0.1915 = 0.3085\)
(B) \(P(190 \le X \le 230)\)
This matches (IV) 0.6247. The two values lie on opposite sides of the mean, so the areas add:
\(P(-0.5 \le Z \le 1.5) = 0.1915 + 0.4332 = 0.6247\)
(C) \(P(X \ge 230)\)
This matches (I) 0.0668.
\(P(Z \ge 1.5) = 0.5 - 0.4332 = 0.0668\)
(D) \(P(210 \le X \le 220)\)
This matches (II) 0.1498. Both values lie above the mean, so the areas subtract:
\(P(0.5 \le Z \le 1) = 0.3413 - 0.1915 = 0.1498\)