\( X \) is a normally distributed random variable with mean 80 and standard deviation 6 . What is the probability that \( X \) is between 70 and 93 ? Write your answer as a decimal rounded to the nearest thousandth.
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To find the probability that \( X \) is between 70 and 93 for a normally distributed random variable with mean \( \mu = 80 \) and standard deviation \( \sigma = 6 \), we can use the Z-score formula: \[ Z = \frac{(X - \mu)}{\sigma} \] 1. Calculate the Z-scores for 70 and 93: For \( X = 70 \): \[ Z_{70} = \frac{(70 - 80)}{6} = \frac{-10}{6} \approx -1.67 \] For \( X = 93 \): \[ Z_{93} = \frac{(93 - 80)}{6} = \frac{13}{6} \approx 2.17 \] 2. Next, we can look up these Z-scores in the standard normal distribution table or use a calculator: - The probability \( P(Z < -1.67) \approx 0.0478 \) - The probability \( P(Z < 2.17) \approx 0.9857 \) 3. Finally, to find the probability that \( X \) is between 70 and 93, we subtract the two probabilities: \[ P(70 < X < 93) = P(Z < 2.17) - P(Z < -1.67) \approx 0.9857 - 0.0478 = 0.9379 \] Thus, the probability that \( X \) is between 70 and 93 is approximately \( \boxed{0.938} \).
