\( X \) is a normally distributed random variable with mean 57 and standard deviation 9. What is the probability that \( X \) is between 48 and 84 ? Use the \( 0.68-0.95-0.997 \) rule and write your answer as a decimal. Round to the nearest thousandth if necessary.
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Extra Insights
To solve this problem, we can utilize the \( 0.68-0.95-0.997 \) rule, which helps us understand the distribution of data in a normal curve based on standard deviations. 1. Calculate the z-scores for the values 48 and 84. - For \( x = 48 \): \[ z = \frac{48 - 57}{9} = \frac{-9}{9} = -1 \] - For \( x = 84 \): \[ z = \frac{84 - 57}{9} = \frac{27}{9} = 3 \] 2. According to the \( 0.68-0.95-0.997 \) rule: - Approximately 68% of the data lies within 1 standard deviation (i.e., between \( 48 \) and \( 66 \)). - Approximately 95% of the data lies within 2 standard deviations (i.e., between \( 39 \) and \( 75 \)). - Approximately 99.7% of the data lies within 3 standard deviations (i.e., between \( 30 \) and \( 84 \)). Since 48 and 84 are within the range of 1 to 3 standard deviations, we can find that \( X \) being between 48 and 84 covers almost the full extent of \( 99.7\% \) of the distribution: \[ P(48 < X < 84) \approx 0.997 \] Thus, rounding to the nearest thousandth, the probability that \( X \) is between 48 and 84 is \( \text{0.997} \).