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)

**i) Proof:**<br /><br />We are given that *d_j = x_j - A*. We want to prove *x̄ = A + d̄*.<br /><br />By definition, the weighted mean *x̄* is:<br /><br />*x̄ = (Σf_jx_j) / n*<br /><br />Substitute *x_j = d_j + A*:<br /><br />*x̄ = (Σf_j(d_j + A)) / n*<br /><br />Distribute the summation:<br /><br />*x̄ = (Σf_jd_j + Σf_jA) / n*<br /><br />Since *A* is a constant, we can factor it out of the summation:<br /><br />*x̄ = (Σf_jd_j + AΣf_j) / n*<br /><br />We know that *Σf_j = n*, so:<br /><br />*x̄ = (Σf_jd_j) / n + (An) / n*<br /><br />*x̄ = (Σf_jd_j) / n + A*<br /><br />By definition, *d̄ = (Σf_jd_j) / n*, therefore:<br /><br />*x̄ = d̄ + A*<br /><br /><br />**ii) Proof:**<br /><br />We want to prove *s² = (Σf_jd_j² - nd̄²) / (n - 1)*.<br /><br />By definition, the variance *s²* is:<br /><br />*s² = (Σf_j(x_j - x̄)²) / (n - 1)*<br /><br />Substitute *x_j = d_j + A* and *x̄ = d̄ + A*:<br /><br />*s² = (Σf_j((d_j + A) - (d̄ + A))²) / (n - 1)*<br /><br />Simplify inside the parentheses:<br /><br />*s² = (Σf_j(d_j - d̄)²) / (n - 1)*<br /><br />Expand the square:<br /><br />*s² = (Σf_j(d_j² - 2d_jd̄ + d̄²)) / (n - 1)*<br /><br />Distribute the summation:<br /><br />*s² = (Σf_jd_j² - 2d̄Σf_jd_j + d̄²Σf_j) / (n - 1)*<br /><br />We know that *Σf_j = n* and *d̄ = (Σf_jd_j) / n*, so *Σf_jd_j = nd̄*:<br /><br />*s² = (Σf_jd_j² - 2d̄(nd̄) + d̄²n) / (n - 1)*<br /><br />Simplify:<br /><br />*s² = (Σf_jd_j² - 2nd̄² + nd̄²) / (n - 1)*<br /><br />*s² = (Σf_jd_j² - nd̄²) / (n - 1)*<br /><br /><br />Therefore, both parts of the question are proven.<br />