The question is asking for the probability that a financial variable, X, which is assumed to follow a normal distribution with a mean (μ) of 40 and a standard deviation (σ) of 14, falls outside the range of 12 to 61.

To solve this, we first convert the given values of X into z-scores using the formula $Z = \frac{X - \mu}{\sigma}$.

For the lower bound (12): $Z = \frac{12 - 40}{14} = -2$

For the upper bound (61): $Z = \frac{61 - 40}{14} = 1.5$

Next, we look up these z-scores in a standard normal distribution table (Z-table) to find the probabilities associated with them.

The probability that Z is less than -2 (P(Z < -2)) is 0.0228. This represents the probability that X is less than 12.

The probability that Z is greater than 1.5 (P(Z > 1.5)) is 0.0668. This represents the probability that X is greater than 61.

To find the combined probability that X is either less than 12 or greater than 61, we add these two probabilities together:

$P(Z < -2) \text{ or } P(Z > 1.5) = P(Z < -2) + P(Z > 1.5) = 0.0228 + 0.0668 = 0.0896$

This means there is an 8.96% chance that the variable X will be outside the range of 12 to 61, which corresponds to option C.