The physical plant at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. the distribution of the number of daily requests is bell-shaped and has a mean of 63 and a standard deviation of 6. using the 68-95-99.7 rule, what is the approximate percentage of light bulb replacement requests numbering between 63 and 75? % (round your answer to the nearest tenth of a percent)

the approximate percentage of light bulb replacement requests between 63 and 75 is 47.5%

Step-by-step explanation:

the 68-95-99.7 rule for a bell-shaped distribution states that

68% of the population is found between mean+ 1 standard deviation and the mean - 1 standard deviation

95% of the population is found between mean+ 2 standard deviations and the mean - 2 standard deviations

99.7% of the population is found between mean+ 3 standard deviations and the mean - 3 standard deviations

therefore since the

light bulbs = mean + x * standard deviation

75 = 63 + x* 6

x= (75-63)/6 = 2

therefore

mean + 2 * standard deviation = 75

mean - 2 standard deviations = 63- 2*6 = 51

and

95% of the light bulbs replacement requests are found between 51 and 75 daily requests

since the bell shaped distribution is symmetrical

95%/2 = 47.5 % is found between mean and mean + 2 * standard deviation

thus

47.5 % of the light bulbs replacements are found between 63 and mean + 2 * standard deviation = 75

47.5 % of the light bulbs replacements are found 51 and 63

18.36. give me brainliest!

step-by-step explanation:

you have to multiply 1.53 by 12 so the equation is 12x1.53 and then that is your answer.