\( X \) is a normally distributed random variable with mean 14 and standard deviation 25 . What is the probability that \( X \) is between 9 and 68 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that \( X \) is between 9 and 68 for a normally distributed random variable with mean 14 and standard deviation 25, we can use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] where \( \mu \) is the mean, and \( \sigma \) is the standard deviation. 1. Calculate the Z-scores for 9 and 68: - For \( X = 9 \): \[ Z_1 = \frac{9 - 14}{25} = \frac{-5}{25} = -0.2 \] - For \( X = 68 \): \[ Z_2 = \frac{68 - 14}{25} = \frac{54}{25} = 2.16 \] 2. Now we look up these Z-scores in a standard normal distribution table or use a calculator to find the corresponding probabilities: - The probability for \( Z_1 = -0.2 \) is approximately \( 0.4207 \) (or 42.07%). - The probability for \( Z_2 = 2.16 \) is approximately \( 0.9846 \) (or 98.46%). 3. To find the probability that \( X \) is between 9 and 68, we subtract the probability of \( Z_1 \) from the probability of \( Z_2 \): \[ P(9 < X < 68) = P(Z < 2.16) - P(Z < -0.2) = 0.9846 - 0.4207 = 0.5639 \] Rounding to the nearest thousandth, the probability that \( X \) is between 9 and 68 is: \[ \text{Probability} \approx 0.564 \]