If $x$ is a normal variate with mean 12 and standard deviation 4, then $P[X ≥ 20]$ is:

[Given that: $P[0 ≤ Z ≤2] = 0.4772$]

0.0228

The correct answer is Option (2) → 0.0228

Given: $X \sim N(\mu=12, \sigma=4),\ P[X \ge 20] = ?$

Standardize: $Z = \frac{X-\mu}{\sigma} = \frac{20-12}{4} = 2$

Then $P[X \ge 20] = P[Z \ge 2] = 1 - P[Z \le 2]$

Using symmetry: $P[Z \ge 2] = P[Z \le -2]$

Given $P[0 \le Z \le 2] = 0.4772$ and $P[Z \le 0] = 0.5$, so $P[Z \le 2] = 0.5 + 0.4772 = 0.9772$

Therefore $P[X \ge 20] = 1 - 0.9772 = 0.0228$