National 5 Maths
Statistics: Data Sets

To use the first formula, we need to calculate the mean.

Then we make a table with these headings:

$$ \begin{array} {|c|c|} \hline x & x-\overline x & (x-\overline x)^2 \\ \hline \sf\small 6 & \sf\small 6-8=-2 & \sf\small (-2)^2=4 \\ \hline \sf\small 8 & \sf\small 8-8=0 & \sf\small 0^2=0 \\ \hline \sf\small 11 & \sf\small 11-8=3 & \sf\small 3^2=9 \\ \hline \sf\small 12 & \sf\small 12-8=4 & \sf\small 4^2=16 \\ \hline \sf\small 8 & \sf\small 8-8=0 & \sf\small 0^2=0 \\ \hline \sf\small 5 & \sf\small 5-8=-3 & \sf\small (-3)^2=9 \\ \hline \sf\small 9 & \sf\small 9-8=1 & \sf\small 1^2=1 \\ \hline \sf\small 7 & \sf\small 7-8=-1 & \sf\small (-1)^2=1 \\ \hline \sf\small 6 & \sf\small 6-8=-2 & \sf\small (-2)^2=4 \\ \hline & \Sigma{\left( x-\overline x \right) }=0 & \Sigma{\left( x-\overline x \right)^2 }= 44 \\ \hline \end{array} $$

\( \Sigma{\left( x-\overline x \right)} \) isn't used in the formula but it should always be zero, so it's a good idea to check before squaring each entry in the middle column. If it doesn't add to zero, go and find your mistake!

We are finally ready to finish this off:

$$ \begin{eqnarray} s &=& \sqrt{\frac{\Sigma{\left( x-\overline x \right)^2 }}{n-1}} \\[6pt] &=& \sqrt{\frac{44}{9-1}} \\[6pt] &=& \sqrt{\frac{44}{8}} \\[6pt] &=& 2.34521... \\[6pt] &\approx& 2.3^\circ C \end{eqnarray} $$

The second formula doesn't need the mean. We make a different table:

$$ \begin{array} {|c|c|} \hline x & x^2 \\ \hline \sf\small 7 & \sf\small 7^2=49 \\ \hline \sf\small 6 & \sf\small 6^2=36 \\ \hline \sf\small 10 & \sf\small 10^2=100 \\ \hline \sf\small 10 & \sf\small 10^2=100 \\ \hline \sf\small 9 & \sf\small 9^2=81 \\ \hline \sf\small 6 & \sf\small 6^2=36 \\ \hline \sf\small 8 & \sf\small 8^2=64 \\ \hline \sf\small 7 & \sf\small 7^2=49 \\ \hline \sf\small 7 & \sf\small 7^2=49 \\ \hline \Sigma{x}=70 & \Sigma{x^2}=564 \\ \hline \end{array} $$

Then we substitute into the formula:

$$ \begin{eqnarray} s &=& \sqrt{\frac{\Sigma{x^2}-\frac{\left(\Sigma x \right)^2}{n}}{n-1}} \\[6pt] &=& \sqrt{\frac{564-\frac{70^2}{9}}{9-1}} \\[6pt] &=& \sqrt{\frac{564-4900\!\div\!9}{8}} \\[6pt] &=& 1.56347... \\[6pt] &\approx& 1.6^\circ C \end{eqnarray} $$

First we'll need to quickly calculate the mean for Edinburgh:

\( \overline x = \large \frac{\Sigma x}{n} \normalsize = \large \frac{70}{9} \normalsize \approx 7.8^\circ C \)

It can be a good idea to put the means and standard deviations into a comparison table:

Comparing the means, we can write any of the following:

• On average, Glasgow was warmer.
• In general, Glasgow was warmer.
• On average, Glasgow had higher temperatures.
• In general, Glasgow had higher temperatures.
• On average, Edinburgh was colder.
• In general, Edinburgh was colder.
• On average, Edinburgh had lower temperatures.
• In general, Edinburgh had lower temperatures.

Comparing the standard deviations, we can write any of the following:

• Glasgow's temperatures were more varied.
• Edinburgh's temperatures were less varied.
• Glasgow's temperatures were less consistent.
• Edinburgh's temperatures were more consistent.
• The spread of temperatures was more in Glasgow.
• The spread of temperatures was less in Edinburgh.

Either formula could be used here, although the arithmetic would be slightly tricker if we used the second version. We'll use the first.

Then we make our table:

$$ \begin{array} {|c|c|} \hline x & x-\overline x & (x-\overline x)^2 \\ \hline \sf\small 1 & \sf\small 1-3=-2 & \sf\small (-2)^2=4 \\ \hline \sf\small 2 & \sf\small 2-3=-1 & \sf\small (-1)^2=1 \\ \hline \sf\small 2 & \sf\small 2-3=-1 & \sf\small (-1)^2=1 \\ \hline \sf\small 2 & \sf\small 2-3=-1 & \sf\small (-1)^2=1 \\ \hline \sf\small 8 & \sf\small 8-3=5 & \sf\small 5^2=25 \\ \hline & \Sigma{\left( x-\overline x \right) }=0 & \Sigma{\left( x-\overline x \right)^2 }= 32 \\ \hline \end{array} $$

Finally, we calculate the standard deviation:

$$ \begin{eqnarray} s &=& \sqrt{\frac{\Sigma{\left( x-\overline x \right)^2 }}{n-1}} \\[6pt] &=& \sqrt{\frac{32}{5-1}} \\[6pt] &=& \sqrt{\frac{32}{4}} \\[6pt] &=& \sqrt{8} \end{eqnarray} $$

The question says that the standard deviation equals \(\sqrt{a}.\)  So \(a=8.\)

(a) First, let's find the mean:

We could use either formula for standard deviation here, but as we've already found the mean, let's choose the first version.

$$ \begin{array} {|c|c|} \hline x & x-\overline x & (x-\overline x)^2 \\ \hline \sf\small 32 & \sf\small 32-31=1 & \sf\small 1^2=1 \\ \hline \sf\small 27 & \sf\small 27-31=-4 & \sf\small (-4)^2=16 \\ \hline \sf\small 34 & \sf\small 34-31=3 & \sf\small 3^2=9 \\ \hline \sf\small 29 & \sf\small 29-31=-2 & \sf\small (-2)^2=4 \\ \hline \sf\small 31 & \sf\small 31-31=0 & \sf\small 0^2=0 \\ \hline \sf\small 33 & \sf\small 33-31=2 & \sf\small 2^2=4 \\ \hline & \Sigma{\left( x-\overline x \right) }=0 & \Sigma{\left( x-\overline x \right)^2 }= 34 \\ \hline \end{array} $$

$$ \begin{eqnarray} s &=& \sqrt{\frac{\Sigma{\left( x-\overline x \right)^2 }}{n-1}} \\[6pt] &=& \sqrt{\frac{34}{6-1}} \\[6pt] &=& \sqrt{\frac{34}{5}} \\[6pt] &=& 2.6076... \\[6pt] &\approx& 2.6 \end{eqnarray} $$

(b) It can be a good idea to put the means and standard deviations into a comparison table:

On average, the Monday buses had more passengers, as 31 > 28.
The number of passengers on each bus varied more on Saturday, as 3.2 > 2.6.

(a) In ascending order, the ten ages are:
31 33 35 36 38 41 41 42 47 55

The median in a sample of ten numbers is the mean of the 5th and 6th, so:

Median \(= \large\frac{38+41}{2}\normalsize = \large\frac{79}{2}\normalsize = 39.5\) years.

\(Q_1 = 35\) and \(Q_3 = 42\)
So \(IQR = 42-35 = 7\) years.

(b) On average, the readers of the newspaper are older, as 41 > 39.5.
The ages of the newspaper readers are more varied, as 9 > 7.

(a) The prices are already in ascending order.

The median in a sample of six numbers is the mean of the 3rd and 4th, so:

Median \(= \large\frac{190+210}{2}\normalsize = \large\frac{400}{2}\normalsize =\) £\(200\small.\)

\(Q_1 = 160\) and \(Q_3 = 230\)
So \(IQR = 230-160 =\) £\(70\small.\)

(b) On average, the cameras are cheaper on the website, as 195 < 200.
The camera prices are more varied on the website, as 73 > 70.