\( X \) is a normally distributed random variable with mean 49 and standard deviation 12 . What is the probability that \( X \) is less than \( 47 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To determine the probability that \( X \) is less than 47, where \( X \) is normally distributed with a mean (\( \mu \)) of 49 and a standard deviation (\( \sigma \)) of 12, follow these steps: 1. **Standardize the Value:** Convert \( X = 47 \) to a standard normal variable \( Z \) using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{47 - 49}{12} = \frac{-2}{12} = -0.1667 \] 2. **Use the Standard Normal Distribution:** We need to find \( P(Z < -0.1667) \). This corresponds to the cumulative distribution function (CDF) of the standard normal distribution at \( Z = -0.1667 \). 3. **Approximate the CDF:** Using standard normal distribution tables or a calculator, we find: \[ P(Z < -0.1667) \approx 0.434 \] This is achieved through either interpolation between standard table values or using a Taylor series expansion for higher precision. 4. **Final Answer:** The probability that \( X \) is less than 47 is approximately **0.434** when rounded to the nearest thousandth. **Answer:** 0.434