Question

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 60 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 27 and 60? the approximate percentage of lightbulb replacement requests numbering between 27 and 60 is %.(do not enter the percent symbol.) question help: video submit question

Explanation:

Step1: Calculate the number of standard deviations from the mean

The mean \(\mu = 60\) and the standard deviation \(\sigma=11\). Let's find \(k\) for \(x = 27\) using the formula \(x=\mu - k\sigma\).

$$27=60 - k\times11$$

$$k\times11=60 - 27$$

$$k\times11 = 33$$

$$k = 3$$

Step2: Use the Empirical Rule

The Empirical Rule states that for a bell - shaped (normal) distribution:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation of the mean (\(\mu\pm\sigma\))
• Approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean (\(\mu\pm2\sigma\))
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean (\(\mu\pm3\sigma\))

The total percentage of data within \(3\) standard deviations of the mean (\(\mu - 3\sigma\) to \(\mu+3\sigma\)) is \(99.7\%\). Since the normal distribution is symmetric about the mean \(\mu\), the percentage of data from \(\mu - 3\sigma\) to \(\mu\) is \(\frac{99.7\%}{2}\)

Answer:

\(49.85\)