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Q1 (1 mark): In lectures we saw that the ‘total deviation about the mean’ is always zero, i.e. ∑= − = n i i x x 1 ( ) 0 . Expand the summation across the brackets and apply this result to the i y to prove that: ( )( ) ( ) 1 1 x x y y x y y i n i i i n i ∑ i − − = ∑ − = = . Ans: Q2 (1 mark) Prove that � (xi − x�)2 𝑛 𝑖=1 = � xi (xi − x�) 𝑛 𝑖=1 Ans:  Q3 (2 m ...

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