\( X \) is a normally distributed random variable with mean 74 and standard deviation 22 . What is the probability that \( X \) is between 8 and 96 ? 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 a normally distributed random variable \( X \) with mean \( \mu = 74 \) and standard deviation \( \sigma = 22 \) falls between 8 and 96, we'll use the **68-95-99.7 (Empirical) Rule**. ### Step 1: Standardize the Interval First, convert the values 8 and 96 to their corresponding **z-scores**: \[ z = \frac{X - \mu}{\sigma} \] - For \( X = 8 \): \[ z = \frac{8 - 74}{22} = \frac{-66}{22} = -3 \] - For \( X = 96 \): \[ z = \frac{96 - 74}{22} = \frac{22}{22} = 1 \] So, we're looking for \( P(-3 \leq Z \leq 1) \). ### Step 2: Apply the Empirical Rule The Empirical Rule states: - **68%** of data lies within **±1σ**, - **95%** within **±2σ**, - **99.7%** within **±3σ**. Using this: - \( P(-3 \leq Z \leq 3) = 99.7\% \) - \( P(-1 \leq Z \leq 1) = 68\% \) - \( P(Z \leq -3) = 0.15\% \) (since half of the remaining 0.3% lies below -3σ) Now, to find \( P(-3 \leq Z \leq 1) \): \[ P(-3 \leq Z \leq 1) = P(Z \leq 1) - P(Z < -3) \] \[ P(Z \leq 1) = 84\% \quad \text{(50\% below mean + 34\% within +1σ)} \] \[ P(Z < -3) = 0.15\% \] \[ P(-3 \leq Z \leq 1) = 84\% - 0.15\% = 83.85\% \] ### Step 3: Convert to Decimal \[ 83.85\% = 0.8385 \] Rounded to the nearest thousandth: \[ 0.839 \] **Final Answer:** The probability is approximately **0.839**.