Mathematics 9709/62 — May/June 2012

Walkthrough

We are given summary data for 150 ages: the total $\sum x = 645$ and the sum of the squares $\sum x^2 = 8287.5$. We want the total squared distance of each data point from the mean, called the sum of squared deviations, $\sum(x - \bar{x})^2$.

Step 1 — Find the mean. The mean is the total divided by the number of values:

$$\bar{x} = \frac{645}{150} = 4.3$$

so the mean age is $4.3$ years.

Step 2 — Use the identity. The sum of squared deviations is linked to the summary statistics by

$$\sum(x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n}$$

This identity comes from expanding the square:

$$\sum(x - \bar{x})^2 = \sum(x^2 - 2x\bar{x} + \bar{x}^2) = \sum x^2 - 2\bar{x}\sum x + n\bar{x}^2$$

and then substituting $\bar{x} = \frac{\sum x}{n}$, which simplifies $2\bar{x}\sum x - n\bar{x}^2$ to $\frac{(\sum x)^2}{n}$.

Step 3 — Substitute the values. Using $\sum x = 645$, $\sum x^2 = 8287.5$ and $n = 150$:

$$\sum(x - \bar{x})^2 = 8287.5 - \frac{645^2}{150} = 8287.5 - 2773.5 = 5514$$

Equivalently, compute the variance first:

$$s^2 = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{8287.5}{150} - 4.3^2 = 55.25 - 18.49 = 36.76$$

and then multiply by $n = 150$:

$$150 \times 36.76 = 5514$$

Key Takeaways

This question shows that the total spread around the mean can be found directly from the two summary statistics $\sum x$ and $\sum x^2$ — no raw data is needed. The central identity

$$\sum(x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n}$$

is the same idea behind the variance formula $s^2 = \frac{\sum x^2}{n} - \bar{x}^2$ and the standard deviation $s = \sqrt{s^2}$. Recognising these equivalent forms and switching between them quickly is a key examination skill.

Common Mistakes

• Forgetting to divide by $n$ when computing the mean: the mean is $\bar{x} = \frac{\sum x}{n}$, not $\sum x$ on its own.
• Using the wrong identity, e.g. writing $\sum x^2 - (\sum x)^2$ instead of $\sum x^2 - \frac{(\sum x)^2}{n}$. The division by $n$ is essential.
• Stopping at the variance: the question asks for $\sum(x - \bar{x})^2$, which is the variance multiplied by $n = 150$. Failing to multiply by 150 loses the final mark (the mark scheme explicitly awards a method mark for the multiplication).
• Rounding intermediate values too aggressively, which can push the final answer outside the accepted range. The mark scheme accepts answers rounding to 5510, but exact working is safest.

Things to Be Careful About

• The mean $4.3$ is exact here because $645 \div 150 = 4.3$; use the exact value rather than a rounded one.
• When using the standard-deviation route, keep $s = 6.063$ to enough significant figures before multiplying by 150, so that $6.063^2 \approx 36.76$ retains accuracy.
• The result has units of squared years, $\text{years}^2$, since it is a sum of squared deviations.
• A faster equivalent route is to skip the standard deviation and evaluate $\sum(x - \bar{x})^2 = 8287.5 - \frac{645^2}{150}$ directly; both routes are acceptable and earn full marks.