\( X \) is a normally distributed random variable with mean 25 and standard deviation 17 . What is the probability that \( X \) is between 8 and 76 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
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To find the probability that \( X \) is between 8 and 76 for a normally distributed variable with a mean of 25 and a standard deviation of 17, we can use the rule of thumb that states: - About 68% of values lie within 1 standard deviation of the mean. - About 95% lie within 2 standard deviations. - About 99.7% lie within 3 standard deviations. First, let's determine how many standard deviations away from the mean the values 8 and 76 are: - For \( X = 8 \): \[ z_1 = \frac{8 - 25}{17} = \frac{-17}{17} = -1 \] - For \( X = 76 \): \[ z_2 = \frac{76 - 25}{17} = \frac{51}{17} \approx 3 \] Now we can interpret these z-scores using the empirical rule. The range between \( z_1 = -1 \) and \( z_2 = 3 \) covers: - 68% for the area between \( z = -1 \) and \( z = 1 \). - 27% from \( z = 1 \) to \( z = 3 \) (approximately, as the area beyond \( z = 3 \) tends towards 0). Adding these two areas gives us: \[ P(8 < X < 76) \approx 0.68 + 0.27 = 0.95 \] When rounded to the nearest thousandth, the probability that \( X \) is between 8 and 76 is: \[ \boxed{0.950} \]
