Year 10A Mean and Standard Deviation Worksheets & Practice

Worked examples

Year 10A mean and standard deviation: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Calculating mean and population standard deviation

   Calculate the mean and population standard deviation of the following five values: 4, 7, 7, 9, 8. Give the standard deviation correct to two decimal places.

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   Key idea: The population standard deviation is $\sigma = \sqrt{\dfrac{\sum(x - \bar{x})^2}{n}}$

   1. Find the mean: $\bar{x} = \frac{4+7+7+9+8}{5} = \frac{35}{5} = 7$

   2. Find each squared deviation from the mean:
      $\begin{aligned}(4-7)^2 &= 9\\(7-7)^2 &= 0\\(7-7)^2 &= 0\\(9-7)^2 &= 4\\(8-7)^2 &= 1\end{aligned}$

   3. Find the population variance: $\sigma^2 = \frac{9+0+0+4+1}{5} = \frac{14}{5} = 2.8$

   4. Find the population standard deviation: $\sigma = \sqrt{2.8} \approx 1.67$

   Answer: Mean $= 7$, population standard deviation $\approx 1.67$

2. Example 2 · Effect of an outlier on standard deviation

   Five students recorded how many hours they exercised in a week: 3, 4, 4, 5, 4. A sixth student joins the group and reports 18 hours of exercise. Calculate the population standard deviation of the original five values and then of all six values. Give each answer correct to two decimal places and describe the effect of the sixth value.

   Show worked solution

   Key idea: An outlier pulls the mean away from the rest of the data, increasing every squared deviation and therefore greatly increasing the standard deviation.

   1. Original five values: mean $= \dfrac{3+4+4+5+4}{5} = \dfrac{20}{5} = 4$

   2. Squared deviations: $(3-4)^2=1,\;(4-4)^2=0,\;(4-4)^2=0,\;(5-4)^2=1,\;(4-4)^2=0$
      $\sigma_5 = \sqrt{\frac{1+0+0+1+0}{5}} = \sqrt{\frac{2}{5}} = \sqrt{0.4} \approx 0.63$

   3. All six values: mean $= \dfrac{20+18}{6} = \dfrac{38}{6} = \dfrac{19}{3} \approx 6.\overline{3}$

   4. Squared deviations from $\dfrac{19}{3}$:
      $\begin{gathered}\begin{aligned}\left(3-\tfrac{19}{3}\right)^2 &= \left(-\tfrac{10}{3}\right)^2 = \tfrac{100}{9}\\\left(4-\tfrac{19}{3}\right)^2 &= \left(-\tfrac{7}{3}\right)^2 = \tfrac{49}{9}\\\left(4-\tfrac{19}{3}\right)^2 &= \tfrac{49}{9}\\\left(5-\tfrac{19}{3}\right)^2 &= \left(-\tfrac{4}{3}\right)^2 = \tfrac{16}{9}\\\left(4-\tfrac{19}{3}\right)^2 &= \tfrac{49}{9}\\\left(18-\tfrac{19}{3}\right)^2 &= \left(\tfrac{35}{3}\right)^2 = \tfrac{1225}{9}\end{aligned} \\ \sigma_6 = \sqrt{\frac{\tfrac{100+49+49+16+49+1225}{9}}{6}} = \sqrt{\frac{1488}{54}} = \sqrt{\frac{248}{9}} \approx \sqrt{27.\overline{5}} \approx 5.25\end{gathered}$

   Answer: $\sigma_5 \approx 0.63$, $\sigma_6 \approx 5.25$

3. Example 3 · Comparing spread using standard deviation

   Two small factories each record their daily output (in units) over five days. Factory A: 20, 22, 21, 23, 24. Factory B: 10, 18, 22, 28, 32. Calculate the population standard deviation for each factory correct to two decimal places and explain which factory has more consistent output.

   Show worked solution

   Key idea: A smaller standard deviation means the data is more tightly clustered around the mean, so output is more consistent.

   1. Factory A mean: $\bar{x}_A = \dfrac{20+22+21+23+24}{5} = \dfrac{110}{5} = 22$

   2. Factory A squared deviations: $(20-22)^2=4,\;(22-22)^2=0,\;(21-22)^2=1,\;(23-22)^2=1,\;(24-22)^2=4$
      $\sigma_A = \sqrt{\frac{4+0+1+1+4}{5}} = \sqrt{\frac{10}{5}} = \sqrt{2} \approx 1.41$

   3. Factory B mean: $\bar{x}_B = \dfrac{10+18+22+28+32}{5} = \dfrac{110}{5} = 22$

   4. Factory B squared deviations: $(10-22)^2=144,\;(18-22)^2=16,\;(22-22)^2=0,\;(28-22)^2=36,\;(32-22)^2=100$
      $\sigma_B = \sqrt{\frac{144+16+0+36+100}{5}} = \sqrt{\frac{296}{5}} = \sqrt{59.2} \approx 7.69$

   5. Both factories have the same mean (22 units), but $\sigma_A \approx 1.41$ is much smaller than $\sigma_B \approx 7.69$, so Factory A has considerably more consistent daily output.

   Answer: $\sigma_A \approx 1.41$, $\sigma_B \approx 7.69$; Factory A is more consistent.

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