Question A.1: Let a be the arithmetic average of a1, a2,…, an. Then (i) ......
Let \bar{a} be the arithmetic average of a_1, a_2,…, a_n. Then
(i) \sum_{i=1}^{n}(a_{i}-{\bar{{a}}})=0,
(i) \sum_{i=1}^{n}(a_{i}-{\bar{a}})^{2}=\sum_{i=1}^{n}a_{i}^{2}-n{\bar{a}}^{2},
We put
S=\sum_{i=1}^{n}a_{i},
so that
{\overline{{a}}}={\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}}={\frac{S}{n}}.Therefore,
\sum_{i=1}^{n}(a_{i}-{\bar{a}})=\sum_{i=1}^{n}a_{i}-\sum_{i=1}^{n}{\bar{a}} (from Proposition 1.1)
=S-n{\bar{a}} (from (1.4))
={S-n\cdot{\frac{S}{n}}}=0.P(A ∪ B) = P(A) + P(B). (1.4)
For Part (ii), we observe first that (put b_{i}=-\overline{{{a}}}\,\,\mathrm{for}\,\,\,\,i=1,2,\ldots,n in the result of Exercise 8)
\sum_{i=1}^{n}(a_{i}-{\overline{{a}}})^{2}=\sum_{i=1}^{n}a_{i}^{2}-2\sum_{i=1}^{n}a_{i}{\overline{{a}}}+\sum_{i=1}^{n}{\overline{{a}}}^{2}.But from the second part of Proposition 1.1 and (1.4) we obtain respectively that
\sum_{i=1}^{n}(a_{i}\bar{a})=\bar{a}\sum_{i=1}^{n}a_{i}=\frac{S}{n}\cdot S=\frac{S^{2}}{n}and
\sum_{n=1}^{n}\overline{{{a}}}^{2}=n\overline{{{a}}}^{2}=n\biggl(\frac{S}{n}\biggr)^{2}=\frac{S^{2}}{n}.Thus, we finally obtain
\sum_{i=1}^{n}\left(a_{i}-{\bar{a}}\right)^{2}=\sum_{i=1}^{n}a_{i}^{2}-2{\frac{S^{2}}{n}}+{\frac{S^{2}}{n}}=\sum_{i=1}^{n}a_{i}^{2}-{\frac{S^{2}}{n}}=\sum_{i=1}^{n}a_{i}^{2}-n\Bigl({\frac{S}{n}}\Bigr)^{2}=\sum_{i=1}^{n}a_{i}^{2}-n{\overline{{a}}}^{2},as required.
Related Answered Questions
Question: A.2
Verified Answer:
Multiplying each term of S by 𝑤 we obtain that
[la...