Let's calculate Standard Deviation


The Standard Deviation for old sequence with mean 15.4 & 10 elements is :

\(SD = \sqrt{( (15.4-10)^2+(15.4-11)^2+(15.4-12)^2+(15.4-15)^2+(15.4-15)^2+(15.4-15)^2+(15.4-17)^2+(15.4-19)^2+(15.4-20)^2+(15.4-20)^2 )/10}\)

The Standard Deviation for new sequence with mean 15.36 with 11 elements is :

\(SD = \sqrt{( (15.36-10)^2+(15.36-11)^2+(15.36-12)^2+(15.36-15)^2+(15.36-15)^2+(15.36-15)^2+(15.36-17)^2+(15.36-19)^2+(15.36-20)^2+(15.36-20)^2 )/11}\)


I think you are right Standard Deviations of both the sequences are not the same

So based upon your logic if you add a number to the sequence that is farther away from mean then it would increase the Standard Deviation value.

yasir, you are partially correct, if you add numbers more than 1 standard deviation from the mean, then SD will increase, if you add numbers less than
1 SD from mean, then SD will decrease, and if you add numbers equal to mean, then SD will remain same.

@Yasir

yes you are right.

adding values closer to the mean ---> decreases SD
adding values farther away from mean ---> increases SD.

@phoenixio

you are wrong...adding a value which is equal to the mean of set, decreases the standard deviation of that set the most.

@Sonalika

You are right, I meant to write in the last sentence, if we add numbers which are at a distance from mean equal to standard deviation, then SD remains same.

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