\( X \) is a normally distributed random variable with mean 89 and standard deviation 4 . What is the probability that \( X \) is between 87 and 91 ? Write your answer as a decimal rounded to the nearest thousandth.
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To find the probability that \( X \) is between 87 and 91, we first need to standardize these values using the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \] Where \( \mu \) is the mean and \( \sigma \) is the standard deviation. 1. For \( X = 87 \): \[ Z_{87} = \frac{87 - 89}{4} = \frac{-2}{4} = -0.5 \] 2. For \( X = 91 \): \[ Z_{91} = \frac{91 - 89}{4} = \frac{2}{4} = 0.5 \] Next, we can find the probabilities corresponding to these Z-scores using the standard normal distribution table or a calculator: - For \( Z = -0.5 \), the probability \( P(Z < -0.5) \) is approximately 0.3085. - For \( Z = 0.5 \), the probability \( P(Z < 0.5) \) is approximately 0.6915. Now, to find the probability that \( X \) is between 87 and 91, we calculate: \[ P(87 < X < 91) = P(Z < 0.5) - P(Z < -0.5) \] \[ = 0.6915 - 0.3085 = 0.3830 \] Thus, the probability that \( X \) is between 87 and 91 is approximately 0.383 (when rounded to the nearest thousandth).
