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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 51 and a standard deviation of 10. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 51 and 61? Do not enter the percent symbol. ans = 95 Incorrect %

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Answer:

The probability that the light bulb replacement requests numbering is between 51 and 61 is 0.34.

Step-by-step explanation:

Let X = number of daily requests to replace florescent light bulbs.

It is provided that[tex]X\sim N(\mu=51, \sigma^{2} = 10^{2})[/tex]

The 68-95-99.7 rule states that:

  • About 68% of the observations fall with 1 standard deviation of mean, i.e.[tex]P(\mu-\sigma<X<\mu+\sigma)=P(41<X<61)=0.68[/tex]
  • About 95% of the observations fall within 2 standard deviations of mean, i.e. [tex]P(\mu-2\sigma<X<\mu+2\sigma)=P(31<X<71)=0.95[/tex]
  • About 99.7% of the observations fall within 3 standard deviations of mean, i.e. [tex]P(\mu-3\sigma<X<\mu+3\sigma)=P(21<X<81)=0.997[/tex]  

Compute the probability that the light bulb replacement requests numbering is between 51 and 61 as follows:

[tex]P(51<X<61)=P(\mu<X<\mu+\sigma)\\=P(X<\mu+\sigma)-P(X<mu)\\=(2.5\%+13.5\%+34\%+34\%)-P(2.5\%+13.5\%+34\%)\\=34\%\\=0.34[/tex]

**Consider the diagram below.

Thus, the probability that the light bulb replacement requests numbering is between 51 and 61 is 0.34.

Ver imagen warylucknow

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