\( X \) is a normally distributed random variable with mean 33 and standard deviation 13 . What is the probability that \( X \) is between 18 and 68 ? Write your answer as a decimal rounded to the nearest thousandth.
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To calculate the probability that \( X \) is between 18 and 68 for a normally distributed random variable \( X \) with mean (µ) 33 and standard deviation (σ) 13, we can use the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] First, we calculate the Z-scores for both values: 1. For \( X = 18 \): \[ Z_{18} = \frac{(18 - 33)}{13} = \frac{-15}{13} \approx -1.154 \] 2. For \( X = 68 \): \[ Z_{68} = \frac{(68 - 33)}{13} = \frac{35}{13} \approx 2.692 \] Next, we look up these Z-scores in the standard normal distribution table or use a calculator to find the probabilities. - The probability corresponding to \( Z = -1.154 \) is approximately 0.1251. - The probability corresponding to \( Z = 2.692 \) is approximately 0.9963. Now, to find the probability that \( X \) is between 18 and 68, we calculate: \[ P(18 < X < 68) = P(Z < 2.692) - P(Z < -1.154) = 0.9963 - 0.1251 \approx 0.8712 \] Therefore, the probability that \( X \) is between 18 and 68 is approximately \( 0.871 \) when rounded to the nearest thousandth.
