\( X \) is a normally distributed random variable with mean 16 and standard deviation 25 . What is the probability that \( X \) is between 41 and 91 ? 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 41 and 91 in a normally distributed random variable, we first standardize these values using the mean (16) and standard deviation (25). 1. Calculate the z-scores for 41 and 91: \[ z_{41} = \frac{41 - 16}{25} = 1 \quad \text{and} \quad z_{91} = \frac{91 - 16}{25} = 3 \] 2. According to the \( 0.68 - 0.95 - 0.997 \) rule: - About 68% of the data lies within 1 standard deviation from the mean (between \( \mu - \sigma \) to \( \mu + \sigma \), which is \( 16 - 25 \) to \( 16 + 25 \) or -9 to 41). - About 95% lies within 2 standard deviations (between \( \mu - 2\sigma \) to \( \mu + 2\sigma \), which is \( 16 - 50 \) to \( 16 + 50 \) or -34 to 66). - About 99.7% lies within 3 standard deviations (between \( \mu - 3\sigma \) to \( \mu + 3\sigma \), which is \( 16 - 75 \) to \( 16 + 75 \) or -59 to 91). Since we want the probability of \( X \) being between 41 and 91, we can see this spans from +1σ (41) up to +3σ (91). Thus, the area/probability between these z-scores is: \[ P(41 < X < 91) = P(Z < 3) - P(Z < 1) \] According to the 0.68-0.95-0.997 rule: - For Z ≤ 1: ~0.8413 (or about 84.13%) - For Z ≤ 3: ~0.9987 (or about 99.87%) Now, calculating the difference: \[ P(41 < X < 91) \approx 0.9987 - 0.8413 = 0.1574 \] In decimal form, the probability that \( X \) is between 41 and 91 is approximately **0.157** when rounded to the nearest thousandth.