Given that ∑x=35 and n=5.
∴¯x=∑xn=355=7.
Let us find ∑x2
Now, ∑(x−9)2=82
⇒∑(x2−18x+81)=82
⇒∑x2−(18∑x)+(81∑1)=82
⇒∑x2−630+405=82 ∵∑x=35 and ∑1=5
⇒∑x2=307.
To find ∑(x−¯x)2, let us consider
∑(x−9)2=82
⇒∑(x−7−2)2=82
⇒∑[(x−7)−2]2=82
⇒∑(x−7)2−2∑[(x−7)×2]+∑4=82
⇒∑(x−¯x)2−4∑(x−¯x)+4∑1=82
⇒∑(x−¯x)2−4(0)+(4×5)=82 ∵∑1=5 and ∑(x−¯x)=0
⇒∑(x−¯x)2=62
∴∑x2=307 and ∑(x−¯x)2=62.