Given:
i=1∑n(xi+1)2=9n
⇒i=1∑n(xi2+2xi+1)=9n...(1)
And, i=1∑n(xi−1)2=5n
⇒i=1∑n(xi2−2xi+1)=5n...(2)
Form (1)+(2), we get
i=1∑n(2xi2+2)=14n
⇒i=1∑nxi2=6n...(3)
From (1)−(2), we get
⇒i=1∑nxi=n...(4)
∴ Variance, σ2=E(x2)−(E(x))2
⇒σ2=n∑x2−n∑x
⇒σ2=n6n−(nn)2=6−1=5
∴ Standard deviation, σ=5.