Porisma

Worked solution

Use raw and shifted summary totals

Find unbiased estimates of the population mean and variance for each sample:

(a) ∑x=120\sum x=120, ∑x2=2102\sum x^2=2102, n=8n=8.

(b) ∑(t−300)=2012\sum(t-300)=2012, ∑(t−300)2=525262\sum(t-300)^2=525262, n=200n=200.

CJC · Lecture · Example 4

(a) Unshifted summaries

xˉ=120/8=15\bar x=120/8=15 and s2=2102−1202/87=3027≈43.1.\begin{aligned}&s^2\\ & =\frac{2102-120^2/8}{7}\\ & =\frac{302}{7}\approx43.1.\end{aligned}

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(b) Shifted summaries

The mean must restore the shift: tˉ=2012/200+300=310.06\bar t=2012/200+300=310.06. A shift does not change variance, so s2=525262−20122/200199=126255324975≈2540(3 s.f.).s^2=\frac{525262-2012^2/200}{199}=\frac{12625532}{4975}\approx2540\quad(3\text{ s.f.}). The exact value is about 2537.795377. Do not add 300 to the variance.

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Common mistakes

Use the sum of squared shifted observations, not the square of their sum; both occur in the formula.

Solution notes

Student has working spaces; Teacher fills both calculations. No distinct lecturer box or printed marks on this page.

Read the hint