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anonymous

  • one year ago

MEDAL!! A certain company recorded the number of employee absences each week over a period of 10 weeks. The result is the data list 3,5,1,2,4,7,4,5,5. Find the mean and standard deviation of the number of absences per week.

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  1. anonymous
    • one year ago
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    I already know the mean is 3.8

  2. anonymous
    • one year ago
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    soooo whats you thinkin|dw:1443620024113:dw|

  3. anonymous
    • one year ago
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    |dw:1443620050104:dw|

  4. carlyleukhardt
    • one year ago
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    Mean days is 4.1 Standard Deviation is 1.81

  5. anonymous
    • one year ago
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    @carlyleukhardt can you explain how you got that please?

  6. carlyleukhardt
    • one year ago
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    sure hang on lemme type it all >.<

  7. anonymous
    • one year ago
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    thanks I appreciate it! I got my other question like this right I just can't figure out how i'm getting this wrong lol

  8. carlyleukhardt
    • one year ago
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    In order to find variance you must first find the mean. The mean is just the average, so add all your inputs and divide by the total number of inputs. Step 1 find mean. 6+5+1+2+2+4+7+4+5+5=41 divide by total inputs (10) = 4.1 is mean days Variance is done by subtracting the mean from each term then squaring the result. Then add up all those results and divide by number of inputs. (6-4.1)=1.9^2=3.61 (5-4.1)=.9^2=.81 (1-4.1)=-3.1^2=9.61 (2-4.1)=-2.1^2=4.41 (2-4.1)=-2.1^2=4.41 (4-4.1)=-.1^2=.01 (7-4.1)=2.9^2=8.41 (4-4.1)=-1^2=.01 (5-4.1)=.9^2=.81 (5-4.1)=.9^2=.81 Now add them up and divide by 10 again. 3.61+.81+9.61+4.41+4.41+.01+8.41+.01+.81+.81/10=3.29 Standard deviation=sqrt of the variance = (sqrt3.29)=1.81

  9. anonymous
    • one year ago
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    thank you so much!

  10. anonymous
    • one year ago
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    you put a 6 it should be a 3! haha that's why you didn't get 38..3.8 like me

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