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nignag [31]
3 years ago
15

The physical plant at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. Th

e distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. Using the empirical rule, what is the approximate percentage of lightbulb replacement requests numbering between 38 and 56?
Mathematics
1 answer:
ankoles [38]3 years ago
6 0

We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.

First of all, we will find z-score corresponding to 38 and 56.

z=\frac{x-\mu}{\sigma}

z=\frac{38-38}{6}=\frac{0}{6}=0

Now we will find z-score corresponding to 56.

z=\frac{56-38}{6}=\frac{18}{6}=3

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is -3\sigma\text{ to }3\sigma.

We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.

We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

\frac{99.7\%}{2}=49.85\%

Therefore, approximately 49.85\% of lightbulb replacement requests numbering between 38 and 56.

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