MATH SOLVE

4 months ago

Q:
# 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)

Accepted Solution

A:

Answer: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 deviation95% of the population is found between mean+ 2 standard deviations and the mean - 2 standard deviations99.7% of the population is found between mean+ 3 standard deviations and the mean - 3 standard deviationstherefore since the light bulbs = mean + x * standard deviation75 = 63 + x* 6x= (75-63)/6 = 2thereforemean + 2 * standard deviation = 75mean - 2 standard deviations = 63- 2*6 = 51and 95% of the light bulbs replacement requests are found between 51 and 75 daily requestssince the bell shaped distribution is symmetrical95%/2 = 47.5 % is found between mean and mean + 2 * standard deviationthus47.5 % of the light bulbs replacements are found between 63 and mean + 2 * standard deviation = 7547.5 % of the light bulbs replacements are found 51 and 63