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.

Question

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.

anonymous · Answer

I already know the mean is 3.8

anonymous · Answer

soooo whats you thinkin|dw:1443620024113:dw|

anonymous · Answer

|dw:1443620050104:dw|

carlyleukhardt · Answer

Mean days is 4.1
 
Standard Deviation is 1.81

anonymous · Answer

@carlyleukhardt can you explain how you got that please?

carlyleukhardt · Answer

sure hang on lemme type it all >.<

anonymous · Answer

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

carlyleukhardt · Answer

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

anonymous · Answer

thank you so much!

anonymous · Answer

you put a 6 it should be a 3! haha that's why you didn't get 38..3.8 like me