\( X \) is a normally distributed random variable with mean 48 and standard deviation 23 . What is the probability that \( X \) is between 2 and 71 ? 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 2 and 71 in a normal distribution with mean \( \mu = 48 \) and standard deviation \( \sigma = 23 \), we can first standardize the values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] For \( X = 2 \): \[ z_1 = \frac{(2 - 48)}{23} \approx -2 \] For \( X = 71 \): \[ z_2 = \frac{(71 - 48)}{23} \approx 1 \] Now we refer to the empirical rule: Approximately 68% of values lie within 1 standard deviation, 95% within 2 standard deviations, and 99.7% within 3 standard deviations of the mean. Since \( z_1 \) corresponds to roughly the -2 standard deviation mark and \( z_2 \) to the 1 standard deviation mark, the probability that \( X \) falls between these two z-scores can be calculated by determining the area under the normal curve from \( z = -2 \) to \( z = 1 \). This area represents: 1. Approximated by the empirical rule: About 95% of values lie within 2 standard deviations (\( -2 \) to \( 2 \)). 2. Since \( -2 \) to \( 0 \) is about 47.5% and \( 0 \) to \( 1 \) is about 34%, we're adding those portions: - From \( -2 \) to \( 0 \): 47.5% - From \( 0 \) to \( 1 \): 34% Adding these portions gives us: \[ P(2 < X < 71) \approx 47.5\% + 34\% = 81.5\% \] So as a decimal rounded to the nearest thousandth, the probability that \( X \) is between 2 and 71 is: \[ \text{Probability} \approx 0.815 \]