Answer: 8.96%

The question is asking for the probability that a random variable \( X \), which follows a normal distribution with a mean \( \mu \) of 40 and a standard deviation \( \sigma \) of 14, does not lie between the values 12 and 61. To find this probability, we first calculate the standardized \( Z \)-scores for the given values using the formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 12 \), the \( Z \)-score is: \[ Z = \frac{12 - 40}{14} = -2 \] For \( X = 61 \), the \( Z \)-score is: \[ Z = \frac{61 - 40}{14} = 1.5 \] Using a standard normal distribution table (Z-table), we find the probabilities for \( Z \) being less than -2 and greater than 1.5. The probability \( P(Z < -2) \) is approximately 0.0228, and \( P(Z > 1.5) \) is approximately 0.0668. Since we are looking for the probability that \( X \) does not lie between 12 and 61, we are interested in the combined probability of \( Z \) being less than -2 or greater than 1.5. The combined probability is calculated by adding the individual probabilities: \[ P(Z < -2) + P(Z > 1.5) = 0.0228 + 0.0668 = 0.0896 \] This gives us the probability that \( X \) does not lie between 12 and 61 as 8.96%, which corresponds to option C. This question tests the understanding of how to work with probabilities in the context of a normal distribution and the use of Z-tables to find probabilities associated with specific \( Z \)-scores.

Author: LeetQuiz Editorial Team