End of chapter exercises

\(140-12=128\)

128 is 1 standard deviation to the left of the mean, therefore the percentage of teenagers who sent less than 128 messages is:

\(\text{50}\% - \text{34}\% = \text{16}\%\)

116 minutes is 2 standard deviations from the mean, therefore \(\text{47,5}\%\)

152 minutes is 1 standard deviation from the mean, therefore \(\text{34}\%\)

Percentage of the teenagers who sent between 116 and 152 messages \(= \text{47,5}\% + \text{34}\% = \text{81,5}\%\)

\[\hat{y} = \text{86 893,33} + \text{3 497,14}x\]

\begin{align*} \text{Profit seventh month } &= \text{86 893,33} + \text{3 497,14}(7) = \text{R}\,\text{111 373,31} \\ \text{Profit eighth month } &= \text{86 893,33} + \text{3 497,14}(8) = \text{R}\,\text{114 870,45} \end{align*}

\begin{align*} \text{130 000} &= \text{86 893,33} + \text{3 497,14}x \\ \therefore x &= \frac{\text{130 000}-\text{86 893,33}}{\text{3 497,14}} = \text{12,33} \end{align*}

It will take 13 months to reach a profit of \(\text{R}\,\text{130 000}\).

\(\hat{y}= \text{601,28} + \text{3,59}x\)

\(r = \text{0,86}\)

There is a strong, positive, linear correlation.

\begin{align*} \text{2 500} &= \text{601,28} + \text{3,59}x \\ \therefore x&= \frac{\text{2 500} - \text{601,28}}{\text{3,59}} = \text{528,89} \end{align*}

Therefore 528 burgers are produced.

\[y= \text{601,28} + \text{3,59}(490) = \text{R}\,\text{2 360,38}\]

\(\hat{y}=\text{77,32} + \text{3,47}x\)

\(r=\text{0,85}\)

The length of the right biceps and the height of the students have a strong, positive linear relationship.

Using a calculator, \(r=\text{0,98}\) which is a very strong, positive, linear correlation between the marks of the first and the second test.

Using a calculator, the least squares regression line equation is:

\[\hat{y} = \text{1,08} + \text{0,57}x\]

Therefore, the expected mark for the second test for the learner in Row 18 is:

\[y = \text{1,08} + \text{0,57}(17) = \text{10,77}\]

Therefore the expected mark for the learner in row 18 for the second test is 11 out of 30.

Average no. of units (\(x\))	Peak temp. (\(y\))	\(xy\)	\(x^{2}\)
37	32	\(\text{1 184}\)	\(\text{1 369}\)
45	40	\(\text{1 800}\)	\(\text{2 025}\)
35	30	\(\text{1 050}\)	1225
30	28	840	900
20	25	500	400
40	38	\(\text{1 520}\)	\(\text{1 600}\)
38	36	\(\text{1 368}\)	\(\text{1 444}\)
15	20	\(\text{300}\)	\(\text{225}\)
20	24	\(\text{480}\)	\(\text{400}\)
22	26	\(\text{572}\)	\(\text{484}\)
\(\sum = 302\)	\(\sum = 299\)	\(\sum = \text{9 614}\)	\(\sum = \text{10 072}\)

\begin{align*} b & = \frac{n{\sum }_{i=1}^{n}{x}_{i}{y}_{i}-{\sum }_{i=1}^{n}{x}_{i}{\sum }_{i=1}^{n}{y}_{i}}{n{\sum }_{i=1}^{n}{\left({x}_{i}\right)}^{2}-{\left({\sum }_{i=1}^{n}{x}_{i}\right)}^{2}} \\ & = \frac{10 \times \text{9 614} - 302 \times \text{299}}{10 \times \text{10 072} - 302^{2}} = \text{0,613913409} \\ \\ a&= \bar{y}-b\bar{x} = \frac{\text{299}}{\text{10}} - \text{0,613913409} \times \frac{302}{10} = \text{11,359815048} \\ \\ \therefore \hat{y}&= \text{11,36} + \text{0,61}x \end{align*}

We have already calculated the value of \(b\) by hand in the question above, so we are left to determine \(\sigma_{x}\) and \(\sigma_{y}\).

Average no. of units (\(x\))	Peak temp. (\(y\))	\((x-\bar{x})^{2}\)	\((y-\bar{y})^{2}\)
32	37	\(\text{46,24}\)	\(\text{4,41}\)
40	45	\(\text{219,04}\)	\(\text{102,01}\)
30	35	\(\text{0,01}\)	\(\text{23,04}\)
28	30	\(\text{0,04}\)	\(\text{3,61}\)
25	20	\(\text{104,04}\)	\(\text{24,01}\)
38	40	\(\text{96,04}\)	\(\text{65,61}\)
36	38	\(\text{60,84}\)	\(\text{37,21}\)
20	15	\(\text{231,04}\)	\(\text{98,01}\)
24	20	\(\text{104,04}\)	\(\text{34,81}\)
26	22	\(\text{67,24}\)	\(\text{15,21}\)
\(\sum=299\)	\(\sum=\text{302}\)	\(\sum=\text{951,6}\)	\(\sum=\text{384,9}\)

\begin{align*} \sigma_{x}&= \frac{\sqrt{\sum\limits_{i=1}^{n}(y_i - \bar{y})^{2}}}{n} = \frac{\sqrt{\text{951,6}}}{10} = \pm \text{3,08} \\ b&=\text{1,52} \\ \sigma_{y}&= \frac{\sqrt{\sum\limits_{i=1}^{n}(x_i - \bar{x})^{2}}}{n} = \frac{\sqrt{\text{384,9}}}{10} = \pm \text{1,96} \\ \therefore r&= \text{0,61} \times \frac{\text{3,08}}{\text{1,96}} \\ &= \text{0,96} \end{align*}

There is a very strong, positive, linear correlation between peak temperature and the average number of electricity units a household uses.

\begin{align*} 45&= \text{11,36} + \text{0,61}x \\ \therefore x &=\frac{45-\text{11,36}}{\text{0,61}} \\ &= \text{55,15} \approx \text{55}\text{ units} \end{align*}

The value we were asked to predict is outside the range of the available data. This is known as extrapolation.

A quadratic function would best fit the data. At about \(\text{18}\)\(\text{°C}\) average household electricity usage is at its minimum. As the peak temperature gets colder or warmer than this point, electrcity usage increases.

This question requires us to find the minimum value of the quadratic equation. There are a number of ways to do this, two are shown below:

The first method is using the formula \(x = \frac{-b}{2a}\):

• The first step is to write the equation in the form: \(y=ax^{2} + bx + c\). Our equation is already in this form, so we can immediately substitute the values into the formula for \(x\). \[x = \frac{-b}{2a} = \frac{\text{4,3}}{(2 \times \text{0,13})} = \text{16,54}\]
• To find \(y\), we substitute our \(x\)-value into the quadratic equation: \(\text{0,13}(\text{16,54})^2 - \text{4,3}(\text{16,54}) + 45 = \text{9,44}\)

Another method is using differentiation:

• The first step is to write the equation in the form: \(y=ax^{2} + bx + c\). Our equation is already in this form, so we can immediately differentiate the equation. \[y' = \text{0,13}(2)x - \text{4,3} = \text{0,26}x + \text{4,3}\]
• At the turning point, \(y' = 0\), therefore we can now solve for \(x\): \begin{align*} 0&= \text{0,26}x - \text{4,3} \\ \therefore x &= \frac{\text{4,3}}{\text{0,26}} = \text{16,54} \end{align*}
• The \(x\)-value can now be substituted into the quadratic equation to find \(y\): \[y = \text{0,13}(\text{16,54})^{2} - \text{4,3}(\text{16,54}) + 45 = \text{9,44}\]

Therefore the peak temperature when electricity demand is at its lowest is \(\text{16,54}\)\(\text{°C}\) and the respective average household electricity usage is \(\text{9,44}\) \(\text{units}\).

The \(y\)-intercept is approximately 1000. At \(x=4\), \(y\) is approximately \(\text{11 000}\). Therefore, \(m = \frac{\Delta y}{\Delta x} = \frac{11000-1000}{4-0} = \text{2 500}\)

The equation for the line of best fit: \(y = \text{2 500}x + \text{1 000}\)

\(a = \text{1 133,996106}\) and \(b = \text{2 393,736978}\), therefore \(\hat{y} = \text{1 134,00} + \text{2 393,74}x\)

\(r = \text{0,85}\)

There is a strong, positive, linear correlation between \(\text{CO}_{2}\) emissions per annum and GDP per capita for the countries in the data set.

\begin{align*} \text{1 712} &= \text{1 134,00} + \text{2 393,74}x \\ \therefore x &= \frac{\text{1 712}-\text{1 134,00}}{\text{2 393,74}} = \text{0,24} \text{ tonnes} \end{align*}

\begin{align*} a&= \text{91,27} \\ b &= -\text{4,91} \\ \hat{y} &= \text{91,27} - \text{4,91}x \end{align*}

\(r=-\text{0,87}\)

The greater the number of Saturdays absent, the lower the mark.

\begin{align*} \hat{y} &= \text{91,27} - \text{4,91}(4) \\ &= \text{71,63}\% \\ &\approx \text{72}\% \end{align*}

Both data sets show negative, linear trends. The trend in Grant's data appears to be more rapidly decreasing than the trend in Christie's data.

\begin{align*} \hat{y}_{\text{Grant}}&= \text{126,13} -\text{3,8}x \\ \hat{y}_{\text{Christie}}&= \text{102,4} - \text{1,11}x \end{align*}

\begin{align*} \text{Grant: } r&= -\text{0,98} \quad \text{(negative, very strong)} \\ \text{Christie: } r&= -\text{0,86} \quad \text{(negative, strong)} \end{align*}

Grant will beat Christie when \(\hat{y}_{\text{Grant}} < \hat{y}_{\text{Christie}}\). To find where the trends intersect, we equate each \(\hat{y}\).

\begin{align*} \text{126,13} -\text{3,8}x &= \text{102,4} - \text{1,11}x \\ -\text{3,8}x + \text{1,11}x&= \text{102,4} - \text{126,13} \\ -\text{2,69}x &= -\text{23,73} \\ x &= \text{8,82} \end{align*}

The race takes place in week 8. \(\text{8,82} > 8\), therefore, Grant will be unable to beat Christie's time when the race takes place.

See answer to e). Grant will be able to beat Christie's time in the ninth week.