skewed right

symmetric

skewed left

skewed right

skewed left

The statistics of the data set are

• mean: \(\text{99,9}\);
• first quartile: \(\text{66,5}\);
• median: \(\text{99,5}\);
• third quartile: \(\text{130,5}\).

Note that we get contradicting indications from the different ways of determining whether the data is skewed right or left.

• The mean is slightly greater than the median. This would indicate that the data set is skewed right.
• The median is slightly closer to the third quartile than the first quartile. This would indicate that the data set is skewed left.

Since these differences are so small and since they contradict each other, we conclude that the data set is symmetric.

The mean is \(\bar{x} = \frac{222}{12}=\text{18,5}\).

To determine the five number summary, we order the data:

\[3\ ;\ 6\ ;\ 7\ ;\ 7\ ;\ 9\ ;\ 11\ ;\ 12\ ;\ 16\ ;\ 31\ ;\ 35\ ;\ 40\ ;\ 45\]

We use the diagram below (or the formulae) to find at, or between, which values the quartiles lie.

The minimum is 3 and the maximum is 45.

For the first quartile, the position is between the third and fourth values. Since both these values are equal to \(\text{7}\), the first quartile is \(\text{7}\).

For the median (second quartile), the position is halfway between the sixth and seventh values. The sixth value is \(\text{11}\) and the seventh value is \(\text{12}\), which means that the median is \(\frac{11+12}{2}=\text{11,5}\).

For the third quartile, the position is between the ninth and tenth values. Therefore the third quartile is \(\frac{31+35}{2}=33\).

Therefore, the five number summary is (3; 7; \(\text{11,5}\); 33; 45)

Mean \(>\) median, therefore the data is skewed right.

The mean is \(\bar{x} = \frac{490}{12}=\text{40,83}\).

To determine the five number summary, we order the data:

\[8\ ;\ 16\ ;\ 21\ ;\ 21\ ;\ 27\ ;\ 31\ ;\ 31\ ;\ 33\ ;\ 38\ ;\ 65\ ;\ 99\ ;\ 100\]

We use the formulae (or a diagram) to find at, or between, which values the quartiles lie.

\begin{align*} \text{Location of } Q_{1} & = \frac{1}{4}(n-1)+1 = \frac{1}{4}(12-1)+1 = \text{3,75}\\ \text{Location of } Q_{2} & = \frac{1}{2}(n-1)+1 = \frac{1}{2}(12-1)+1 = \text{6,5}\\ \text{Location of } Q_{3} & = \frac{3}{4}(n-1)+1 = \frac{3}{4}(12-1)+1 = \text{9,25} \end{align*}

The minimum is 8 and the maximum is 100.

For the first quartile, the position is between the third and fourth values. Since both these values are equal to \(\text{21}\), the first quartile is \(\text{21}\).

For the median (second quartile), the position is halfway between the sixth and seventh values. The sixth value and seventh value are \(\text{31}\), which means that the median is \(\text{31}\).

For the third quartile, the position is between the ninth and tenth values. Therefore the third quartile is \(\frac{38+65}{2}=\text{51,5}\).

Therefore, the five number summary is (8; 21; \(\text{31}\); \(\text{51,5}\); 100)

Mean \(>\) median, therefore the data is skewed right.

The mean is \(\bar{x} = \frac{654}{10}=\text{65,4}\).

To determine the five number summary, we order the data:

\[11\ ;\ 46\ ;\ 57\ ;\ 58\ ;\ 65\ ;\ 77\ ;\ 77\ ;\ 81\ ;\ 90\ ;\ 92\]

We use the formulae (or a diagram) to find at, or between, which values the quartiles lie.

\begin{align*} \text{Location of } Q_{1} & = \frac{1}{4}(n-1)+1 = \frac{1}{4}(10-1)+1 = \text{3,25}\\ \text{Location of } Q_{2} & = \frac{1}{2}(n-1)+1 = \frac{1}{2}(10-1)+1 = \text{5,5}\\ \text{Location of } Q_{3} & = \frac{3}{4}(n-1)+1 = \frac{3}{4}(10-1)+1 = \text{7,75} \end{align*}

The minimum is 11 and the maximum is 92.

For the first quartile, the position is between the third and fourth values. Since the third is 57 and the fourth is 58, the first quartile is \(\frac{57+58}{2}=\text{57,5}\).

For the median (second quartile), the position is halfway between the fifth and sixth values. The sixth value is 65 and the seventh value is \(\text{77}\), therefore the median is \(\frac{65+77}{2}=\text{71}\).

For the third quartile, the position is between the seventh and eighth values. Therefore the third quartile is \(\frac{77+81}{2}=\text{79}\).

Therefore, the five number summary is (11; \(\text{57,5}\); \(\text{77}\); \(\text{79}\); 92)

Mean \(<\) median, therefore the data is skewed left.

The mean is \(\bar{x} = \frac{641}{11}=\text{58,27}\).

To determine the five number summary, we order the data:

\[1\ ;\ 17\ ;\ 33\ ;\ 41\ ;\ 50\ ;\ 74\ ;\ 76\ ;\ 76\ ;\ 83\ ;\ 91\ ;\ 99\]

We use the formulae (or a diagram) to find at, or between, which values the quartiles lie.

\begin{align*} \text{Location of } Q_{1} & = \frac{1}{4}(n-1)+1 = \frac{1}{4}(11-1)+1 = \text{3,5}\\ \text{Location of } Q_{2} & = \frac{1}{2}(n-1)+1 = \frac{1}{2}(11-1)+1 = \text{6}\\ \text{Location of } Q_{3} & = \frac{3}{4}(n-1)+1 = \frac{3}{4}(11-1)+1 = \text{8,5} \end{align*}

The minimum is 1 and the maximum is 99.

For the first quartile, the position is between the third and fourth values. The third value is 33 and the fourth value is 41, therefore the first quartile is \(\frac{33+41}{2}=\text{37}\).

The second quartile (median) is at the sixth position. The sixth value is \(\text{74}\).

For the third quartile, the position is between the eighth and ninth values. Therefore the third quartile is \(\frac{76+83}{2}=\text{79,5}\).

Therefore, the five number summary is (1; 37; \(\text{74}\); \(\text{79,5}\); 99)

Mean \(<\) median, therefore the data is skewed left.

The mean is \(\bar{x} = \frac{-\text{7,8}}{13}=-\text{0,6}\).

To determine the five number summary, we order the data:

\[-\text{5,2};-\text{2};-\text{1,8};-\text{1,8};-\text{1,7};-\text{0,9};-\text{0,5};-\text{0,2}\ ;\ \text{0,1}\ ;\ \text{0,5}\ ;\ \text{0,5}\ ;\ \text{2,2}\ ;\ \text{3}\]

We use the formulae (or a diagram) to find at, or between, which values the quartiles lie.

\begin{align*} \text{Location of } Q_{1} & = \frac{1}{4}(n-1)+1 = \frac{1}{4}(13-1)+1 = \text{4}\\ \text{Location of } Q_{2} & = \frac{1}{2}(n-1)+1 = \frac{1}{2}(13-1)+1 = \text{7}\\ \text{Location of } Q_{3} & = \frac{3}{4}(n-1)+1 = \frac{3}{4}(13-1)+1 = \text{10} \end{align*}

The minimum is \(-\text{5,2}\) and the maximum is 3.

The first quartile is at the fourth position. Therefore, the first quartile is \(-\text{1,8}\).

The median (second quartile) is at the seventh position. The seventh value is \(-\text{0,5}\).

The third quartile is at the tenth position. Therefore the third quartile is \(\text{0,5}\).

Therefore, the five number summary is (\(-\text{5,2}\); \(-\text{1,8}\); \(-\text{0,5}\); \(\text{0,5}\); 3)

Mean \(\approx\) median, therefore the data is symmetric.

The mean is \(\bar{x} = \frac{742}{12}=\text{61,83}\).

To determine the five number summary, we order the data:

\[25\ ;\ 31\ ;\ 44\ ;\ 46\ ;\ 54\ ;\ 60\ ;\ 64\ ;\ 71\ ;\ 78\ ;\ 86\ ;\ 86\ ;\ 97\]

We use the formulae (or a diagram) to find at, or between, which values the quartiles lie.

\begin{align*} \text{Location of } Q_{1} & = \frac{1}{4}(n-1)+1 = \frac{1}{4}(12-1)+1 = \text{3,75}\\ \text{Location of } Q_{2} & = \frac{1}{2}(n-1)+1 = \frac{1}{2}(12-1)+1 = \text{6,5}\\ \text{Location of } Q_{3} & = \frac{3}{4}(n-1)+1 = \frac{3}{4}(12-1)+1 = \text{9,25} \end{align*}

The minimum is 25 and the maximum is 97.

For the first quartile, the position is between the third and fourth values. The third value is 44 and the fourth value is \(\text{46}\), which means that the first quartile is \(\frac{44+46}{2}=\text{45}\).

For the median (second quartile), the position is halfway between the sixth and seventh values. The sixth value is 60 and the seventh value is \(\text{64}\), which means that the median is \(\frac{60+64}{2}=\text{62}\).

For the third quartile, the position is between the ninth and tenth values. Therefore the third quartile is \(\frac{78+86}{2}=\text{82}\).

Therefore, the five number summary is (25; 45; \(\text{62}\); \(\text{82}\); 97)

Mean \(\approx\) median, therefore the data is symmetric.

The formula for the mean is

\begin{align*} \bar{x} &= \frac{\sum\limits_{i=1}^n x_i}{n} \\ \therefore \bar{x} &= \frac{\text{47,2}}{8} \\ &= \text{5,90} \end{align*}

The formula for the variance is \[\sigma^2 = \frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}\]

We first subtract the mean from each data point and then square the result.

\(x_i\)	\(\text{9,1}\)	\(\text{0,2}\)	\(\text{2,8}\)	\(\text{2,0}\)	\(\text{10,0}\)	\(\text{5,8}\)	\(\text{9,3}\)	\(\text{8,0}\)
\(x_i - \overline{x}\)	\(\text{3,2}\)	\(-\text{5,7}\)	\(-\text{3,1}\)	\(-\text{3,9}\)	\(\text{4,1}\)	\(-\text{0,1}\)	\(\text{3,4}\)	\(\text{2,1}\)
\((x_i - \overline{x})^2\)	\(\text{10,24}\)	\(\text{32,49}\)	\(\text{9,61}\)	\(\text{15,21}\)	\(\text{16,81}\)	\(\text{0,01}\)	\(\text{11,56}\)	\(\text{4,41}\)

The variance is the sum of the last row in this table divided by \(\text{8}\), so \(\sigma^2 = \frac{\text{47,2}}{8}=\text{12,54}\). The standard deviation is the square root of the variance, therefore \(\sigma=\sqrt{12.54}=\pm \text{3,54}\).

The interval containing all values that are one standard deviation from the mean is \(\left[\text{5,90}-\text{3,54} ; \text{5,90}+\text{3,54}\right] = \left[\text{9,44} ; \text{2,36}\right]\). We are asked how many values are within than one standard deviation from the mean, meaning inside the interval. There are \(\text{5}\) values from the data set within the interval, which is \(\frac{5}{8} \times 100=\text{63}\%\) of the data points.

The formula for the mean is

\begin{align*} \bar{x} &= \frac{\sum\limits_{i=1}^n x_i}{n} \\ \therefore \bar{x} &= \frac{55}{10} \\ &= \text{5,5} \end{align*}

The formula for the variance is \[\sigma^2 = \frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}\]

We first subtract the mean from each data point and then square the result.

\(x_i\)	\(\text{9}\)	\(\text{5}\)	\(\text{1}\)	\(\text{3}\)	\(\text{3}\)	\(\text{5}\)	\(\text{7}\)	\(\text{4}\)	\(\text{10}\)	\(\text{8}\)
\(x_i - \overline{x}\)	\(\text{3,5}\)	\(-\text{0,5}\)	\(-\text{4,5}\)	\(-\text{2,5}\)	\(-\text{2,5}\)	\(-\text{0,5}\)	\(\text{1,5}\)	\(-\text{1,5}\)	\(\text{4,5}\)	\(\text{2,5}\)
\((x_i - \overline{x})^2\)	\(\text{12,25}\)	\(\text{0,25}\)	\(\text{20,25}\)	\(\text{6,25}\)	\(\text{6,25}\)	\(\text{0,25}\)	\(\text{2,25}\)	\(\text{2,25}\)	\(\text{20,25}\)	\(\text{6,25}\)

The variance is the sum of the last row in this table divided by \(\text{10}\), so \(\sigma^2 = \frac{\text{76,5}}{\text{10}} = \text{7,65}\). The standard deviation is the square root of the variance, therefore \(\sigma=\sqrt{\text{7,65}}=\pm \text{2,77}\).

The interval containing all values that are one standard deviation from the mean is \(\left[\text{5,5}-\text{2,77} ; \text{5,5}+\text{2,77}\right] = \left[\text{2,73} ; \text{8,27}\right]\). We are asked how many values are within than one standard deviation from the mean, meaning inside the interval. There are \(\text{7}\) values from the data set within the interval, which is \(\frac{7}{10} \times 100=\text{70}\%\) of the data points.

The formula for the mean is

\begin{align*} \bar{x} &= \frac{\sum\limits_{i=1}^n x_i}{n} \\ \therefore \bar{x} &= \frac{516}{12} \\ &= \text{43} \end{align*}

The formula for the variance is \[\sigma^2 = \frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}\]

We first subtract the mean from each data point and then square the result.

\(x_i\)

\(\text{81}\)

\(\text{22}\)

\(\text{63}\)

\(\text{12}\)

\(\text{100}\)

\(\text{28}\)

\(\text{54}\)

\(\text{26}\)

\(\text{50}\)

\(\text{44}\)

\(\text{4}\)

\(\text{32}\)

\(x_i - \overline{x}\)

\(\text{38}\)

\(-\text{21}\)

\(\text{20}\)

\(-\text{31}\)

\(\text{57}\)

\(-\text{15}\)

\(\text{11}\)

\(-\text{17}\)

\(\text{7}\)

\(\text{1}\)

\(-\text{39}\)

\(-\text{11}\)

\((x_i - \overline{x})^2\)

\(\text{1 444}\)

\(\text{441}\)

\(\text{400}\)

\(\text{961}\)

\(\text{3 249}\)

\(\text{225}\)

\(\text{121}\)

\(\text{289}\)

\(\text{49}\)

\(\text{1}\)

\(\text{1 521}\)

\(\text{121}\)

The variance is the sum of the last row in this table divided by \(\text{12}\), so \(\sigma^2 = \frac{\text{8 822}}{12}=\text{735,17}\). The standard deviation is the square root of the variance, therefore \(\sigma=\sqrt{735.17}=\pm \text{27,11}\).

The interval containing all values that are one standard deviation from the mean is \(\left[\text{43}-\text{27,11} ; \text{43}+\text{27,11}\right] = \left[\text{15,89} ; \text{70,11}\right]\). We are asked how many values are within than one standard deviation from the mean, meaning inside the interval. There are \(\text{8}\) values from the data set within the interval, which is \(\frac{8}{12} \times 100=\text{67}\%\) of the data points.

• Mean = \(\text{5}\)
• \(\sigma^{2} = \text{9}\)
• \(\sigma = \pm \text{3}\)

• Mean = \(\text{8,58}\)
• \(\sigma^{2} = \text{30,91}\)
• \(\sigma = \pm \text{5,56}\)

• Mean = \(\text{3,88}\)
• \(\sigma^{2} = \text{1,47}\)
• \(\sigma = \pm \text{1,21}\)

• Mean = \(-\text{1,66}\)
• \(\sigma^{2} = \text{11,47}\)
• \(\sigma = \pm \text{3,39}\)

• Mean = \(\text{54,07}\)
• \(\sigma^{2} = \text{1 406,07}\)
• \(\sigma = \pm \text{37,50}\)

Supermarket A: \(\text{3,84}\). Supermarket B: \(\text{3,82}\). Supermarket B has the lower mean.

Standard deviation: \[\sigma = \sqrt{\frac{\sum\limits_{i=1}^n \left(x_i - \bar{x}\right)^2}{n}}\]

For Supermarket A: \begin{align*} \sigma &= \sqrt{\frac{\sum\limits_{i=1}^n \left(x_i - \text{3,84}\right)^2}{\text{6}}} \\ &= \sqrt{\frac{\text{0,0882}}{\text{6}}} \\ &= \sqrt{\text{0,0147}} \\ &\approx \pm \text{0,121} \end{align*}

For Supermarket B: \begin{align*} \sigma &= \sqrt{\frac{\sum\limits_{i=1}^n \left(x_i - \text{3,82}\right)^2}{6}} \\ &= \sqrt{\frac{\text{0,20}\dot{3}}{\text{6}}} \\ &= \sqrt{\text{0,033}\dot{8}} \\ &\approx \pm\text{0,184} \end{align*}

The standard deviation of Supermarket A's prices is lower than that of Supermarket B's. That means that Supermarket A has more consistent (less variable) prices than Supermarket B.

\(\overline{x} = \text{47,87}\)

\(\sigma = \pm\text{0,27}\)

The mean is \(\text{47,87}\) and the standard deviation is \(\text{0,56}\). Therefore the interval containing all values that are one standard deviation from the mean is \(\left[\text{47,87}-\text{0,27} ; \text{47,87}+\text{0,27}\right] = \left[\text{47,60} ; \text{48,15}\right]\). We are asked how many values are further than one standard deviation from the mean, meaning outside the interval. There are \(\text{3}\) values from the data set outside the interval.

\(90=72 +2(9)\)

Therefore 90 lies at 2 standard deviations to the right of the mean.

Hence, \(\text{47,5}\%\) of students scored between 72 and 90 marks.

\begin{align*} &45=72-3(9) \\ &\therefore 45 \text{ lies at 3 standard deviations to the left of the mean.} \\ &63= 72 - 9 \\ & \therefore 63 \text{ lies at 1 standard deviation to the left of the mean.} \\ & \text{The area between 1 standard deviation and 3 standard deviations is: } \\ &\text{49,5} - 34 = \text{15,5}\% \\ &\therefore \text{15,5}\% \text{ of } 184 \approx 29 \text{ students scored between 45 and 63 marks.} \end{align*}