\( X \) is a normally distributed random variable with mean 46 and standard deviation 8 . What is the probability that \( X \) is between 62 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 62 and 70 in a normal distribution with a mean of 46 and a standard deviation of 8, we first convert these values into z-scores. The z-score for 62 is calculated as: \[ z = \frac{62 - 46}{8} = 2 \] The z-score for 70 is calculated as: \[ z = \frac{70 - 46}{8} = 3 \] Using the \( 0.68-0.95-0.997 \) rule (the empirical rule), we know that: - Approximately 68% of the data falls within 1 standard deviation (between \( \mu - \sigma \) and \( \mu + \sigma \)), - Around 95% falls within 2 standard deviations, - About 99.7% falls within 3 standard deviations. Since \( z = 2 \) (62) is 2 standard deviations above the mean and \( z = 3 \) (70) is 3 standard deviations above the mean, we can estimate that the probability between these z-scores is: - We approximately have 95% of the data values between -2 and +2. - From \( 2 \) to \( 3 \) standard deviations above the mean corresponds to about 2.5% (half of the 5% that lie outside the ±2 standard deviations). Thus, the area under the curve from \( z = 2 \) to \( z = 3 \) can be estimated as: \[ P(62 < X < 70) \approx 0.025 \text{ since this is the additional area from } z=2 \text{ to } z=3. \] Therefore, rounding to the nearest thousandth, the probability that \( X \) is between 62 and 70 is approximately: \[ \text{Probability} \approx 0.025. \]