\( X \) is a normally distributed random variable with mean 44 and standard deviation 18 . What is the probability that \( X \) is between 8 and 98 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.

To find the probability that \( X \) is between 8 and 98 for a normally distributed random variable with mean \( \mu = 44 \) and standard deviation \( \sigma = 18 \), we can use the properties of the normal distribution. 1. **Calculate the Z-scores** for the values 8 and 98 using the formula: \[ Z = \frac{X - \mu}{\sigma} \] 2. **Calculate the Z-score for \( X = 8 \)**: \[ Z_1 = \frac{8 - 44}{18} = \frac{-36}{18} = -2 \] 3. **Calculate the Z-score for \( X = 98 \)**: \[ Z_2 = \frac{98 - 44}{18} = \frac{54}{18} = 3 \] 4. **Use the Z-scores to find the probabilities**: - The Z-score of -2 corresponds to the left tail of the normal distribution. - The Z-score of 3 corresponds to the right tail of the normal distribution. Using the \( 0.68-0.95-0.997 \) rule: - Approximately 68% of the data falls within 1 standard deviation from the mean. - Approximately 95% of the data falls within 2 standard deviations from the mean. - Approximately 99.7% of the data falls within 3 standard deviations from the mean. 5. **Determine the probabilities**: - The probability that \( Z < -2 \) is approximately \( 0.0228 \) (from Z-tables). - The probability that \( Z < 3 \) is approximately \( 0.9987 \) (from Z-tables). 6. **Calculate the probability that \( X \) is between 8 and 98**: \[ P(8 < X < 98) = P(Z < 3) - P(Z < -2) = 0.9987 - 0.0228 \] Now, let's perform the calculation: \[ P(8 < X < 98) = 0.9987 - 0.0228 = 0.9759 \] Thus, the probability that \( X \) is between 8 and 98 is approximately \( 0.976 \) when rounded to the nearest thousandth. Final answer: \[ \boxed{0.976} \]