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 39 and a standard deviation of 11. Using the 68-95-99.7 (Empirical) Rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 50? The approximate percentage of lightbulb replacement requests numbering between 39 and 50 is

27 + 13x + 15 = (x + 5)( =

step-by-step explanation:

27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( = 27 + 13x + 15 = (x + 5)( =

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.