\( X \) is a normally distributed random variable with mean 48 and standard deviation 12 . What is the probability that \( X \) is between 6 and \( 16 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 6 and 16 for a normally distributed random variable with mean 48 and standard deviation 12, we can first standardize these values using the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] For \( X = 6 \): \[ Z_1 = \frac{(6 - 48)}{12} = \frac{-42}{12} = -3.5 \] For \( X = 16 \): \[ Z_2 = \frac{(16 - 48)}{12} = \frac{-32}{12} = -2.67 \] Now, we need to find the probabilities corresponding to these Z-scores using the standard normal distribution table (or a calculator). - The probability for \( Z = -3.5 \) is approximately 0.0002. - The probability for \( Z = -2.67 \) is approximately 0.0038. Now, to find the probability that \( X \) is between 6 and 16, we subtract the cumulative probabilities: \[ P(6 < X < 16) = P(Z < -2.67) - P(Z < -3.5) \approx 0.0038 - 0.0002 = 0.0036 \] Thus, the probability that \( X \) is between 6 and 16 is approximately **0.004** when rounded to the nearest thousandth.