\( X \) is a normally distributed random variable with mean 49 and standard deviation 20 . What is the probability that \( X \) is between 21 and \( 64 ? \) Write your answer as a decimal rounded to the nearest thousandth.

To solve for the probability that \( X \) is between 21 and 64 in a normally distributed variable with mean 49 and standard deviation 20, we first standardize the values using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] For \( X = 21 \): \[ Z_{21} = \frac{21 - 49}{20} = \frac{-28}{20} = -1.4 \] For \( X = 64 \): \[ Z_{64} = \frac{64 - 49}{20} = \frac{15}{20} = 0.75 \] Next, we look up these Z-scores in the standard normal distribution table or use a calculator: - The probability corresponding to \( Z = -1.4 \) is approximately 0.0808. - The probability corresponding to \( Z = 0.75 \) is approximately 0.7734. To find the probability that \( X \) is between 21 and 64, we subtract the two probabilities: \[ P(21 < X < 64) = P(Z < 0.75) - P(Z < -1.4) \approx 0.7734 - 0.0808 = 0.6926 \] Thus, the probability that \( X \) is between 21 and 64 is approximately \( \boxed{0.693} \) when rounded to the nearest thousandth.