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Add sides 5


Subtract sides 2x


Divide sides by 4


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CHECK :



Thus the solution is correct....
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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.


Now we will find z-score corresponding to 56.

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.

Therefore, approximately
of lightbulb replacement requests numbering between 38 and 56.
I do believe that that point is -2.2 because of the fact that the difference between -1.6 and -2.4 is 0.8, and there are 4 marks in between them. So each mark's value is decreased by -0.2.
1 mile=5280 feet
4 miles=21120 feet
21120×20÷100=4224. As a result, 20% of 4 miles is 4224 feet. Hope it help!
Answer:
51.2545454545
Step-by-step explanation: