11.5 Measures of central tendency and variation, including standard deviation — Practise

Work through these on your own. Every standard deviation below divides by $n$ . State that convention in your working.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Find the mean of $4, 7, 9, 12, 13, 15$ .
Find the median of $3, 5, 8, 10, 12, 14$ .
Find the range of $2, 5, 9, 14$ .
Find the variance of $2, 4, 6, 8, 10$ , using the convention that divides by $n$ .
Find the standard deviation of $2, 4, 6, 8, 10$ , dividing by $n$ .
A data set has $n = 10$ , $\sum x = 60$ and $\sum x^2 = 800$ . Find the standard deviation, dividing by $n$ .
Find the interquartile range of $1, 3, 4, 6, 8, 9, 11, 14$ , using the $(n + 1)$ position method.
Find the upper quartile of $1, 3, 4, 6, 8, 9, 11, 14$ , using the $(n + 1)$ position method.
Find the standard deviation, dividing by $n$ , of $10, 12, 14, 16, 18$ , and then of $10, 12, 14, 16, 18, 30$ .
Find the median and the mean of $5, 6, 6, 7, 100$ .
Two data sets both have mean 17. Set A is $12, 15, 17, 18, 23$ and set B is $5, 15, 17, 19, 29$ . Find the standard deviation of each, dividing by $n$ .
Find the mean and the median of $6, 7, 7, 8, 9, 9, 10, 12$ . State which average you would report, and why.

Answers and concise solutions

1. The mean

Answer: Add the values and divide by the number of values: $\dfrac{60}{6} = 10$ .

2. The median

Answer: There are six values, so the median is the mean of the 3rd and 4th: $\dfrac{8 + 10}{2} = 9$ .

3. The range

Answer: The range is the largest value minus the smallest: $14 - 2 = 12$ .

4. The variance

Answer: The mean is 6. The squared deviations are $16, 4, 0, 4, 16$ , which sum to 40. Dividing by $n = 5$ gives variance $\dfrac{40}{5} = 8$ .

5. The standard deviation

Answer: The variance is 8, so $\sigma = \sqrt{8} = 2\sqrt{2} = 2.83$ (3 s.f.).

6. The standard deviation

Answer: The mean is $\bar{x} = 6$ . Then $\sigma = \sqrt{\dfrac{800}{10} - 6^2} = \sqrt{44} = 2\sqrt{11} = 6.63$ (3 s.f.).

7. The interquartile range

Answer: There are $n = 8$ values. The position of $Q_1$ is $\dfrac{9}{4} = 2.25$ , giving $Q_1 = 3.25$ . The position of $Q_3$ is $\dfrac{27}{4} = 6.75$ , giving $Q_3 = 10.5$ . The IQR is $10.5 - 3.25 = 7.25 = \dfrac{29}{4}$ .

8. The upper quartile

Answer: The position of $Q_3$ is $\dfrac{3(8 + 1)}{4} = 6.75$ , so $Q_3 = 9 + 0.75(11 - 9) = 10.5 = \dfrac{21}{2}$ .

9. The two standard deviations

Answer: The first standard deviation is $\sqrt{8} = 2.83$ (3 s.f.). Adding 30 raises the second to $\dfrac{2\sqrt{95}}{3} = 6.50$ (3 s.f.). One outlier can more than double the standard deviation.

10. The median and the mean

Answer: The median is 6 and the mean is $\dfrac{124}{5} = 24.8$ . The single large value pulls the mean far above the median, so the median describes the typical value better here.

11. The two standard deviations

Answer: Set A has $\sigma = \dfrac{\sqrt{330}}{5} = 3.63$ (3 s.f.). Set B has $\sigma = \dfrac{2\sqrt{370}}{5} = 7.69$ (3 s.f.). The means agree, but set B is more spread out.

12. The mean and the median

Answer: The mean is $\dfrac{68}{8} = 8.5$ and the median is $\dfrac{8 + 9}{2} = 8.5$ . The data are fairly symmetric with no outlier, so either average is suitable. The mean uses every value, so report it.