\( X \) is a normally distributed random variable with mean 6 and standard deviation 14 . What is the probability that \( X \) is between 38 and 90 ? Write your answer as a decimal rounded to the nearest thousandth.

To find the probability that a normally distributed random variable \( X \) with mean \( \mu = 6 \) and standard deviation \( \sigma = 14 \) falls between 38 and 90, we can use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Calculating the Z-scores for the values 38 and 90: 1. For \( X = 38 \): \[ Z_{38} = \frac{38 - 6}{14} = \frac{32}{14} \approx 2.286 \] 2. For \( X = 90 \): \[ Z_{90} = \frac{90 - 6}{14} = \frac{84}{14} = 6 \] Now, we will look up these Z-scores in the standard normal distribution table (or use a calculator): - The probability of \( Z < 2.286 \) is approximately 0.9890. - The probability of \( Z < 6 \) is effectively 1 (since 6 is far beyond typical Z-tables). Next, we find the probability that \( X \) is between 38 and 90: \[ P(38 < X < 90) = P(Z < 6) - P(Z < 2.286) \approx 1 - 0.9890 = 0.0110 \] Therefore, the probability that \( X \) is between 38 and 90 is approximately **0.011** (rounded to the nearest thousandth).