\( X \) is a normally distributed random variable with mean 2 and standard deviation 12 What is the probability that \( X \) is between 27 and 36 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 27 and 36 for a normally distributed variable with a mean of 2 and a standard deviation of 12, we can use the standard normal distribution (Z-score). First, we calculate the Z-scores for 27 and 36: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 27 \): \[ Z_{27} = \frac{27 - 2}{12} = \frac{25}{12} \approx 2.0833 \] For \( X = 36 \): \[ Z_{36} = \frac{36 - 2}{12} = \frac{34}{12} \approx 2.8333 \] Next, we can find the probabilities from the standard normal distribution table or using a calculator: - Probability of \( Z < 2.0833 \) is approximately \( 0.9812 \). - Probability of \( Z < 2.8333 \) is approximately \( 0.9976 \). Now, we subtract the two probabilities to find the probability that \( X \) is between 27 and 36: \[ P(27 < X < 36) = P(Z < 2.8333) - P(Z < 2.0833) \approx 0.9976 - 0.9812 = 0.0164 \] Thus, the probability that \( X \) is between 27 and 36 is approximately \( 0.016 \) when rounded to the nearest thousandth.