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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 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 = __________%

Respuesta :

Answer:

50 percent

Step-by-step explanation:

Mean = 39

[tex]\sigma = 3[/tex]

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

[tex](\mu -1 \sigma, \mu +1\sigma)=(39-3,39+3)=(36,42)[/tex]

95% data falls within two standard deviation of mean

[tex](\mu -2 \sigma, \mu +2\sigma)=(39-2(3),39+2(3))=(33,45)[/tex]

99.7% data falls within 3 standard deviation of mean

[tex](\mu -3 \sigma, \mu +3\sigma)=(39-3(3),39+3(3))=(30,48)[/tex]

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 .

Ver imagen wifilethbridge

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