If for a distributionΣ(x−5)=3,Σ(x−5)2=43 and the total number of items is 18, find the mean and standard deviation.

Given, Σ(x−5)=3 and Σ(x−5)2=43,n=18∴  Mean =5+∑(x−5)18=5+318=5+16=316=5.17, Now, Σ(x−5)2=43⇒Σx2−Σ10x+Σ25=43⇒Σx2−10{Σ(x−5+5)}+18×25=43⇒Σx2−10Σ(x−5)−10Σ5+450=43⇒Σx2−10(3)−10×18×5+450=43⇒Σx2−30−900+450=43⇒Σx2=523 and Σ(x−5)=3⇒Σx−5Σ1=3⇒Σx=3+5×18=93∵σ=Σx2n−Σxn2=52318−93182 =29.06−(5.17)2 =29.06−26.73 =2.33=1.53