Mathematics 9709/61 — October/November 2010

Walkthrough

This question gives the two summary quantities you need directly: the total of the observations, $\sum t$, and the sum of the squared deviations from the mean, $\sum (t - \bar{t})^2$.

1. Find the mean. The mean of a set of $n$ observations is

$$\bar{t} = \frac{\sum t}{n}.$$

Here there are 50 days, so

$$\bar{t} = \frac{910}{50} = 18.2.$$

1. Use the given squared deviations. The standard deviation is the square root of the average squared deviation from the mean:

$$s = \sqrt{\frac{\sum (t - \bar{t})^2}{n}}.$$

The question has already told us that $\sum (t - \bar{t})^2 = 876$, so we do not need to calculate each deviation.

1. Evaluate.

$$s = \sqrt{\frac{876}{50}} = \sqrt{17.52} = 4.19 \quad (3 \text{ s.f.})$$

Key Takeaways

• The mean is found by dividing the total of all observations by the number of observations.
• When a set of data is summarised by $\sum (x - \bar{x})^2$, the standard deviation can be computed directly as $\sqrt{\frac{\sum (x - \bar{x})^2}{n}}$.
• Keep the unsimplified expression, such as $\sqrt{876/50}$, visible in your working; examiners award method marks for seeing it.

Common Mistakes

• Forgetting the square root: the variance is $\frac{876}{50} = 17.52$, but the standard deviation is $\sqrt{17.52} \approx 4.19$.
• Using $n - 1$ instead of $n$: unless told the data is a sample for which $n-1$ is required, Cambridge S1 uses the divisor $n$ for the standard deviation of a set of data.
• Using a rounded mean in further calculations: here the mean is exact at $18.2$, so there is no rounding issue, but if the mean had not been exact, using a rounded value early could change the final answer.
• Not showing the method expression: the mark scheme gives a method mark specifically for the correct unsimplified expression $\sqrt{876/50}$, so write it down.

Things to Be Careful About

• The total $\sum t = 910$ is in degrees; the mean and standard deviation will also be in degrees Celsius.
• The standard deviation must be a non-negative number. If you get a negative value, check whether you forgot to take the square root.
• Round the final standard deviation to a suitable degree of accuracy; the mark scheme accepts $4.19$.