Lance P. answered • 10/03/19
SWAG--UM (Students Will Achieve Greatness & Understanding in Math)
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 61 and a standard deviation of 6. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 61 and 73?
The phrase bell-shaped is a buzzword for implying a "normal distribution" curve. We are given a μ (mean) of 61, which means that this is the center of the bell-curve graph or rather the 50% percentile, which also implies a probability of .5000 on the z-score table. The standard deviation (σ) is 6 and the lightbulb replacement requests start at the mean (61) and go two standard deviations to the right (6+6 =12) yielding a 73.
In essence, this is a two-tailed probability P(61 < x < 73) or P( 0 < z < 2) where you subtract probability (z-score normal cumulative distribution) of the upper bound from lower bound; however, you are asked to use the 68/95/99.7 rule, which states that middle 68% comprises of probability values plus and minus one standard deviation away from the mean, the middle 95% comprises of probability values plus/minus two standard deviations away from the mean, and the middle 99.7% comprises of probability values plus/minus three standard deviations away. (Depending on textbook source and instructor, these values may be slightly different but close).
With that said and for this problem, we care about probability values that is "only" plus two standard deviations away from the mean or half of 95% which is 47.5% and is the answer to your question.
