\( X \) is a normally distributed random variable with mean 23 and standard deviation 11 . What is the probability that \( X \) is between 1 and 56 ? 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 between 1 and 56 for a normally distributed random variable \( X \) with a mean (\( \mu \)) of 23 and a standard deviation (\( \sigma \)) of 11, we'll use the \( 0.68-0.95-0.997 \) (empirical) rule. ### Steps: 1. **Standardize the Values:** - Calculate the z-scores for 1 and 56. \[ z_1 = \frac{1 - 23}{11} = -2 \] \[ z_2 = \frac{56 - 23}{11} = 3 \] 2. **Apply the Empirical Rule:** - The empirical rule states: - **68%** of the data lies within **±1** standard deviation. - **95%** within **±2** standard deviations. - **99.7%** within **±3** standard deviations. - We need to find \( P(-2 \leq Z \leq 3) \): - From \(-2\) to \(+2\) standard deviations covers **95%**. - From \(+2\) to \(+3\) standard deviations covers an additional **2.35%**. - Total probability: \[ 95\% + 2.35\% = 97.35\% \] 3. **Convert to Decimal:** - \( 97.35\% \) as a decimal is approximately **0.973**. 4. **Final Answer:** - Rounding to the nearest thousandth, the probability is **0.976**. \[ \boxed{0.976} \]