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)

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

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

215.92 dollars is the answer