Answer: 8.96%

The question is asking for the probability that a random variable X, which is assumed to follow a normal distribution with a mean (μ) of 40 and a standard deviation (σ) of 14, lies outside the range of 12 to 61. To find this probability, we first calculate the standardized (z) scores for the given values 12 and 61 using the formula: \[ Z = \frac{X - \mu}{\sigma} \] For 12: \[ Z = \frac{12 - 40}{14} = -2 \] For 61: \[ Z = \frac{61 - 40}{14} = 1.5 \] Using a Z-table or standard normal distribution table, we find the probabilities for these z-scores. The probability that a z-score is less than -2 is 0.0228, and the probability that a z-score is greater than 1.5 is 0.0668. Since we are looking for the combined probability of X being less than 12 or greater than 61, we add these two probabilities together: \[ P(Z < -2) + P(Z > 1.5) = 0.0228 + 0.0668 = 0.0896 \] Thus, the probability that X lies outside the range between 12 and 61 is 8.96%, which corresponds to option C. This question tests the understanding of the properties of the normal distribution and the ability to calculate probabilities using z-scores.

Author: LeetQuiz Editorial Team