\( X \) is a normally distributed random variable with mean 29 and standard deviation 4 . What is the probability that \( X \) is between 34 and 36 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 34 and 36 for a normally distributed random variable with mean 29 and standard deviation 4, we can standardize the values using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 34 \): \[ Z_{34} = \frac{34 - 29}{4} = \frac{5}{4} = 1.25 \] For \( X = 36 \): \[ Z_{36} = \frac{36 - 29}{4} = \frac{7}{4} = 1.75 \] Next, we can use the standard normal distribution table (or a calculator) to find the probabilities corresponding to these Z-scores: - The probability of \( Z < 1.25 \) is approximately 0.8944. - The probability of \( Z < 1.75 \) is approximately 0.9599. Now, subtract the two probabilities to find the probability that \( X \) falls between 34 and 36: \[ P(34 < X < 36) = P(Z < 1.75) - P(Z < 1.25) \approx 0.9599 - 0.8944 = 0.0655 \] Rounding this to the nearest thousandth: The probability that \( X \) is between 34 and 36 is approximately **0.066**.