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natulia [17]
3 years ago
14

The physical plant at the main campus of a large state university recieves daily requests to replace florecent lightbulbs. The d

istribution of the number of daily requests is bell-shaped and has a mean of 39 and a standard deviation of 3. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 39 and 48?
Do not enter the percent symbol.
ans = __________%

Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

50 percent

Step-by-step explanation:

Mean = 39

\sigma = 3

Using the 68-95-99.7 rule, 68% of the data falls within first  standard deviation of mean

(\mu -1 \sigma, \mu +1\sigma)=(39-3,39+3)=(36,42)

95% data falls within two standard deviation of mean

(\mu -2 \sigma, \mu +2\sigma)=(39-2(3),39+2(3))=(33,45)

99.7% data falls within 3 standard deviation of mean

(\mu -3 \sigma, \mu +3\sigma)=(39-3(3),39+3(3))=(30,48)

Refer the attached graph

The curve is normally distributed

Now the the percentage of light bulb replacement requests numbering between 39 and 48 = 34%+13.5%+2.5% = 50%

Hence  the approximate percentage of light bulb replacement requests numbering between 39 and 48 is 50 .

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