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 41 and a standard deviation of 9. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 41 and 59?(Round your answer to the nearest tenth of a percent)

Answer:47.5%Step-by-step explanation:First, calculate the difference between mean and  the percentage of lightbulbs requiring replacement in the  problem: 59 – 41 = 18 41 – 41 = 0 Since standard deviation is 9 lightbulbs , for the first case there is a difference of 2 standard deviation and for the second is 0 standard deviation. the 68-95-99.7% rule establish that: 1 standard deviation means a 68% of total area under the normal distribution. On one side is half = 34%. 2 standard deviations means a 95% of total area under the normal distribution. On one side is half = 47.5%. 3 standard deviation means a 99.7% of total area under the normal distribution. On one side is half = 49.85%. 0 standard deviation means no area under the curve The difference between 2 and 0 standard deviation is 47.5– 0 = 47.5% The answer is 47.5% of lightbulbs requiring replacement