If this is adding or subtracting you can get your answer by using the formula part over whole so turning it into a fraction and solving so you would have 1500 over 7% hope this helps
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.
Answer:
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Step-by-step explanation:
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Answer:
You didn't ask a question.
Step-by-step explanation:
Natural numbers are counting numbers. Natural numbers cannot be a fraction with a different numerator and denominator when simplified. Therefore 2/5 is not a natural number.