\( X \) is a normally distributed random variable with mean 11 and standard deviation 25 . What is the probability that \( X \) is between 36 and 61 ? 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 36 and 61, we can first standardize the values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu = 11 \) (mean) and \( \sigma = 25 \) (standard deviation). Calculating the z-scores for 36 and 61: 1. For \( X = 36 \): \[ z = \frac{(36 - 11)}{25} = \frac{25}{25} = 1 \] 2. For \( X = 61 \): \[ z = \frac{(61 - 11)}{25} = \frac{50}{25} = 2 \] Now we look at the standard normal distribution and use the \( 0.68-0.95-0.997 \) rule. This rule tells us that: - About 68% of values lie within 1 standard deviation from the mean, - About 95% lie within 2 standard deviations, - About 99.7% lie within 3 standard deviations. Since we are looking for the probability between \( z = 1 \) and \( z = 2 \): - The probability between \( z = -1 \) and \( z = 1 \) is approximately 0.68 (68%), - The probability between \( z = -2 \) and \( z = 2 \) is approximately 0.95 (95%). To find the probability between \( z = 1 \) and \( z = 2 \), we can consider that the area between \( z = -2 \) and \( z = 2 \) encompasses the area between \( z = -1 \) and \( z = 1 \) as well. Thus, we can calculate: \[ P(1 < Z < 2) = P(-2 < Z < 2) - P(-1 < Z < 1) \] This gives us: \[ P(1 < Z < 2) = 0.95 - 0.68 = 0.27 \] Thus, the probability that \( X \) is between 36 and 61 is \( \approx 0.270 \).