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is this right?
Find the standard deviation to the given data. The manager of a small dry cleaner employs six people. As part of their personnel file, she asked each one to record to the nearest one-tenth of a mile the distance they travel one way from home to work. THe six distances are listed:
17.5,12.5,26.0,32.2,17.7,20.1. Round the result to two decimal places.
so the mean that i got was 21
then I found the square of the number within the list and added the results then i used the formula to find variance.
= (2892.64- 6 (21)^2)/ (6-1)
= i am ended up after calculations with
49.328
then i find the deviation squaring the answer.
sqrt 49.328 approx.= 7.023 and rounding to the two decimal places would be: 7.03 that would be my standard deviation
You found the sample n-1 deviation correctly, however, how did you round a 3 up?
to round up a 3 it should be next to it a 5 or above so...
7.023 rounded to two decimal places would look like 702.3 which would be 702.0 then or am i still wrong...
Find the standard deviation to the given data. The manager of a small dry cleaner employs six people. As part of their personnel file, she asked each one to record to the nearest one-tenth of a mile the distance they travel one way from home to work. THe six distances are listed:
17.5,12.5,26.0,32.2,17.7,20.1. Round the result to two decimal places.
so the mean that i got was 21
then I found the square of the number within the list and added the results then i used the formula to find variance.
= (2892.64- 6 (21)^2)/ (6-1)
= i am ended up after calculations with
49.328
then i find the deviation squaring the answer.
sqrt 49.328 approx.= 7.023 and rounding to the two decimal places would be: 7.03 that would be my standard deviation
You found the sample n-1 deviation correctly, however, how did you round a 3 up?
to round up a 3 it should be next to it a 5 or above so...
7.023 rounded to two decimal places would look like 702.3 which would be 702.0 then or am i still wrong...
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