\( X \) is a normally distributed random variable with mean 44 and standard deviation 13 . What is the probability that \( X \) is between 18 and 70 ? 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 18 and 70, we first need to calculate the z-scores for both values. The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu \) is the mean and \( \sigma \) is the standard deviation. For \( X = 18 \): \[ z_{18} = \frac{(18 - 44)}{13} = \frac{-26}{13} = -2 \] For \( X = 70 \): \[ z_{70} = \frac{(70 - 44)}{13} = \frac{26}{13} = 2 \] Now we have z-scores of -2 and 2. According to the \( 0.68-0.95-0.997 \) rule: - About 68% of the data falls within 1 standard deviation from the mean, - About 95% falls within 2 standard deviations, and - About 99.7% falls within 3 standard deviations. Since -2 and 2 correspond to 2 standard deviations from the mean, we can conclude that the probability that \( X \) is between 18 and 70 is approximately 95%. So the answer is: \[ \text{Probability} \approx 0.950 \]