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The owner of a chain of clothing stores is comparing the monthly profit earned in the past year from four different store locations. She calculated the mean and standard deviation of the monthly profit, in dollars, for each location, as shown in the table.

For which store location does 68% of the data lie between $19,371.18 and $22,295.48?

The owner of a chain of clothing stores is comparing the monthly profit earned in the past year from four different store locations She calculated the mean and class=

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Answer:

Answer choice C, location C

Step-by-step explanation:

If you add or subtract the standard deviation to the monthly profit, then you get $19,371.18 and $22,295.48. This shows that the deviation withholds most of the data given

varshamittal029

The store location for which  [tex]68%[/tex]% of the data lies between $19,371.18 and $22.295.48 is location C

What is standard deviation?

Standard deviation explains the relation of data with mean.

How to find the location of the data?

For normal distributions, 68% of the data lies under one SD from the mean. Two standard deviations is within  95%, and three standard deviations would take up to 99%.

This implies that on taking (mean + SD) and (mean - SD), 68% of data would cover these two numbers.

 Add and subtract SD from the mean to check which location will give $19,371.18 and $22.295.48.

For location A, we have

M-SD=23,124.70-1553.43=21,571.27

M SD = 23,124.70 + 1,553.43 = 24,678.4

For location B, we have

M - SD = 24,842.18 - 1,617.20 = 23,224.98

M + SD = 24, 842.18 + 1,617.20 = 26,459.38

For location C, we have

M - SD = 20,833.33 - 1,462.15 = 19,371.18

M + SD = 20,833.33 + 1,462.15 = 22,295.48

This implies the store location is C

Learn more about standard deviation here:

https://brainly.com/question/475676

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