20,825 because I think you just multiply 245 and 85 and that should be the answer
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}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
![z=\frac{38-38}{6}=\frac{0}{6}=0](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B38-38%7D%7B6%7D%3D%5Cfrac%7B0%7D%7B6%7D%3D0)
Now we will find z-score corresponding to 56.
![z=\frac{56-38}{6}=\frac{18}{6}=3](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B56-38%7D%7B6%7D%3D%5Cfrac%7B18%7D%7B6%7D%3D3)
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
.
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\%](https://tex.z-dn.net/?f=%5Cfrac%7B99.7%5C%25%7D%7B2%7D%3D49.85%5C%25)
Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
Answer:
its either 406.417112299 or 63.3333333333
Step-by-step explanation
so srry