Question
e) Suppose that x_(1),x_(2),L,x_(k) occur with frequencies f_(1),f_(2),L,f_(k) , respectively. Let d_(j)=x_(j)-A denote the deviation of the j"observation x_(j),j=1,2,ldots ,k from a number A.. Prove that i) bar (x)=A+bar (d) (2 Marks) ii) s^2=(sum _(j-1)^kf_(j)d_(j)^2-nbar (d)^2)/(n-1) where n=sum _(j=1)^kf_(j) and bar (d)=(1)/(n)sum _(j=1)^kf_(j)d_(j) (3 Marks)
Solution
4.6
(301 Votes)
Jarvis
Master · Tutor for 5 years
Answer
**i) Proof:**We are given that *dj = xj - A*. We want to prove *x̄ = A + d̄*.By definition, the weighted mean *x̄* is:*x̄ = (Σfjxj) / n*Substitute *xj = dj + A*:*x̄ = (Σfj(dj + A)) / n*Distribute the summation:*x̄ = (Σfjdj + ΣfjA) / n*Since *A* is a constant, we can factor it out of the summation:*x̄ = (Σfjdj + AΣfj) / n*We know that *Σfj = n*, so:*x̄ = (Σfjdj) / n + (An) / n**x̄ = (Σfjdj) / n + A*By definition, *d̄ = (Σfjdj) / n*, therefore:*x̄ = d̄ + A***ii) Proof:**We want to prove *s² = (Σfjdj² - nd̄²) / (n - 1)*.By definition, the variance *s²* is:*s² = (Σfj(xj - x̄)²) / (n - 1)*Substitute *xj = dj + A* and *x̄ = d̄ + A*:*s² = (Σfj((dj + A) - (d̄ + A))²) / (n - 1)*Simplify inside the parentheses:*s² = (Σfj(dj - d̄)²) / (n - 1)*Expand the square:*s² = (Σfj(dj² - 2djd̄ + d̄²)) / (n - 1)*Distribute the summation:*s² = (Σfjdj² - 2d̄Σfjdj + d̄²Σfj) / (n - 1)*We know that *Σfj = n* and *d̄ = (Σfjdj) / n*, so *Σfjdj = nd̄*:*s² = (Σfjdj² - 2d̄(nd̄) + d̄²n) / (n - 1)*Simplify:*s² = (Σfjdj² - 2nd̄² + nd̄²) / (n - 1)**s² = (Σfjdj² - nd̄²) / (n - 1)*Therefore, both parts of the question are proven.