The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 51 and a standard deviation of 10 . Using the \( 68-95-99.7 \) rule, what is the approximate percentage of lightbulb replacement requests numbering between 31 and 51 ? Do not enter the percent symbol. ans =

To find the approximate percentage of lightbulb replacement requests between 31 and 51 using the \( 68-95-99.7 \) rule, we first determine how many standard deviations away from the mean these numbers are. The mean is 51, and the standard deviation is 10. For 31, the calculation is: \( 31 - 51 = -20 \) \( -20 / 10 = -2 \) (this is 2 standard deviations below the mean) Now, for 51, it’s the mean itself: \( 51 - 51 = 0 \) \( 0 / 10 = 0 \) (this is the mean) According to the \( 68-95-99.7 \) rule: - About 68% of the data falls within 1 standard deviation from the mean (between 41 and 61). - About 95% falls within 2 standard deviations (between 31 and 71). Since we need the percentage between 31 and 51, we take half of the 95% (since 31 to the mean is half the range of 31 to 71): - Therefore, it's approximately \( 95% / 2 = 47.5% \). However, since this accounts for the left side up to the mean, we summarize: It’s \( 47.5% \) of the area from 31 to 51. So, the answer is: 47.5