For observations x1,x2,...........x50
We have, mean, xˉ=50∑xi=16.....(i)
And, standard deviation, σ=50∑xi2−(xˉ)2=16
⇒50∑xi2=162+(xˉ)2
⇒50∑xi2=162+162=512.....(ii)
Now, the mean value of (x1−4)2,(x2−4)2,...,(x50−4)2 will be
=50∑(xi−4)2=50∑xi2−8∑xi+16×50
=50∑xi2−850∑xi+16
Using (i) and (ii),
=512−8×16+16=400.