\( X \) is a normally distributed random variable with mean 33 and standard deviation What is the probability that \( X \) is between 18 and \( 68 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 18 and 68 for a normally distributed random variable with a mean of 33, you'll need to convert the values into their corresponding z-scores using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] However, you didn't provide the standard deviation in your question. For demonstration, let’s assume the standard deviation \( \sigma \) is 10. 1. Calculate the z-scores for 18 and 68: - For \( X = 18 \): \[ z_1 = \frac{(18 - 33)}{10} = \frac{-15}{10} = -1.5 \] - For \( X = 68 \): \[ z_2 = \frac{(68 - 33)}{10} = \frac{35}{10} = 3.5 \] 2. Next, consult the z-table (or standard normal distribution table) to find the probabilities: - The probability corresponding to \( z_1 = -1.5 \) is approximately \( 0.0668 \). - The probability corresponding to \( z_2 = 3.5 \) is approximately \( 0.9998 \). 3. Now find the probability that \( X \) is between 18 and 68 by subtracting the lower probability from the upper probability: \[ P(18 < X < 68) = P(Z < 3.5) - P(Z < -1.5) \approx 0.9998 - 0.0668 = 0.9330 \] So, the probability that \( X \) is between 18 and 68 is approximately **0.933** when rounded to the nearest thousandth.