The distribution of the number of daily requests is bell-shaped and has a mean of 46 and a standard deviation of 5. using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 36 and 46?

If the mean = 54, 1 standard deviation (+1σ) above the mean = 54+6 = 60

1 standard deviation (-1σ) below the mean = 54 - 6 = 48

68% of the data of a normal distribution lies between -1σ and +1σ: between 48 - 60 requests

between 54 - 60 requests (between the mean and +1σ) is half of this percentage: 68/2 = 34%

34%

i pretty sure it is

x/3-10=8

x/3=18

x=48