\( X \) is a normally distributed random variable with mean 18 and standard deviation 15 . What is the probability that \( X \) is between 48 and 63 ? 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 \( X \) lies between 48 and 63 in a normally distributed variable with mean \( \mu = 18 \) and standard deviation \( \sigma = 15 \), we first calculate the z-scores for both values. 1. For \( X = 48 \): \[ z = \frac{X - \mu}{\sigma} = \frac{48 - 18}{15} = \frac{30}{15} = 2 \] 2. For \( X = 63 \): \[ z = \frac{X - \mu}{\sigma} = \frac{63 - 18}{15} = \frac{45}{15} = 3 \] Now we want to find the probability that \( X \) is between z-scores of 2 and 3. According to the \( 0.68-0.95-0.997 \) rule, approximately: - About 68% of the data falls within 1 standard deviation (between -1 and 1 z-scores). - About 95% falls within 2 standard deviations (between -2 and 2 z-scores). - About 99.7% falls within 3 standard deviations (between -3 and 3 z-scores). To find the probability between z = 2 and z = 3, we calculate the area under the normal curve for these z-scores: - Probability for \( z = 2 \) (between -2 and 2): 95% (or 0.95). This means \( 0.95 / 2 = 0.475 \) is to the right of the mean. - Probability for \( z = 3 \) (between -3 and 3): 99.7% (or 0.997). This means \( 0.997 / 2 = 0.4985 \) is to that side of the mean. Now, to find the area between z = 2 and z = 3: \[ P(2 < z < 3) = P(z < 3) - P(z < 2) = 0.9987 - 0.975 \] Calculating that gives us: \[ P(48 < X < 63) \approx 0.9987 - 0.975 \approx 0.0237 \] So the probability that \( X \) is between 48 and 63 is approximately **0.024** when rounded to the nearest thousandth.