\( X \) is a normally distributed random variable with mean 34 and standard deviation 3 . What is the probability that \( X \) is less than 43 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To determine the probability that \( X \) is less than 43, given that \( X \) is normally distributed with a mean (\( \mu \)) of 34 and a standard deviation (\( \sigma \)) of 3, we can use the 68-95-99.7 (empirical) rule. ### Steps: 1. **Identify the Number of Standard Deviations:** \[ 43 = \mu + k\sigma \implies 43 = 34 + k \times 3 \implies k = \frac{43 - 34}{3} = 3 \] So, 43 is 3 standard deviations above the mean. 2. **Apply the Empirical Rule:** - **Within 1 standard deviation (\( \mu \pm 1\sigma \)):** Approximately 68% of the data. - **Within 2 standard deviations (\( \mu \pm 2\sigma \)):** Approximately 95% of the data. - **Within 3 standard deviations (\( \mu \pm 3\sigma \)):** Approximately 99.7% of the data. 3. **Interpret the Result:** Since 43 is \( \mu + 3\sigma \), it represents the upper bound of the 99.7% range. This means that approximately 99.7% of the data lies below 43. ### Final Answer: \[ 0.997 \]