DSE 2004 Doubt plz help

[MohitG]

10 outcomes of a RV were recorded. The sample mean is 0 and variance is 4. It is discovered that two entries were recorded incorrectly: one was recorded as -5 instead of -6 while the other was recorded as 11 instead of 12. Calculate the actual variance. Th answer is given to be 7.4. But somehow I just can't figure out this one. Plz help :(

[Granpa Simpson]

Original Mean=0
Corrected Mean=0-{(-5)-11}+{(-6)+11}=0.
Now, original variance=4,
Thus sum(xi^2)=4*10=40.
Now corrected sum(xi^2)=40-(25+121)+(36+144)=74.
Now new variance=1/n*corrected sum(xi^2)-(Corrected Mean)^2.
Since corrected mean=0, new variance=1/n*corrected sum(xi^2)=74/10=7.4

[MohitG]

Thanks a ton Garcia......

[Granpa Simpson]

My pleasure bro..