Question

1.The set of data a={ 12,6,7,3,15,10,18,5 } and b={ 9,3,8,8,9,8,9,18 }.
i. Measure the central tendency(mean, median, mode ,GM,HM) of a given data set and explain which one is the poor measure
ii. Measure the dispersion(range, mean deviation, standard deviation, variance) of a given data set and explain which one is the poor measure
iii. Measure the skewness and kurtosis of a given dataset and explain

Answer 1

Answer on Question #62250 – Math – Statistics and Probability

Question

The set of data $a = \{12,6,7,3,15,10,18,5\}$ and $b = \{9,3,8,8,9,8,9,18\}$.

i. Measure the central tendency (mean, median, mode, GM, HM) of a given data set and explain which one is the poor measure

ii. Measure the dispersion (range, mean deviation, standard deviation, variance) of a given data set and explain which one is the poor measure

iii. Measure the skewness and kurtosis of a given dataset and explain

Solution

i. For $a = \{12,6,7,3,15,10,18,5\}$:

$$mean = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{12 + 6 + 7 + 3 + 15 + 10 + 18 + 5}{8} = 9.5$$

Rearrange elements of the set a as follows:

$$a = \{3, 5, 6, 7, 10, 12, 15, 18\}.$$

The sample size $n = 8$ is even, hence the median is the mean of the two middle values in the sorted list:

$$median = \frac{7 + 10}{2} = 8.5.$$

There is no mode, because there is no value that appears most often in the set of data.

$$GM = \sqrt[n]{x_1 x_2 x_3 \dots x_n} = \sqrt[8]{12 \cdot 6 \cdot 7 \cdot 3 \cdot 15 \cdot 10 \cdot 18 \cdot 5} = 8.2$$ $$HM = \frac{n}{\sum_{i=1}^{n} \frac{1}{x_i}} = \frac{8}{\frac{1}{12} + \frac{1}{6} + \frac{1}{7} + \frac{1}{3} + \frac{1}{15} + \frac{1}{10} + \frac{1}{18} + \frac{1}{5}} = 7.0$$

We do not have the mode for this data. So, the mode is the poor measure of central tendency for this data.

For $b = \{9,3,8,8,9,8,9,18\}$:

$$mean = \frac{\sum_{i=1}^{m} y_i}{m} = \frac{9 + 3 + 8 + 8 + 9 + 8 + 9 + 18}{8} = 9$$

Rearrange elements of the set b as follows:

$$b = \{3, 8, 8, 8, 9, 9, 9, 18\}.$$

The sample size $m = 8$ is even, hence the median is the mean of the two middle values in the sorted list:

$$median = \frac{8 + 9}{2} = 8.5.$$

There are two modes: 8 and 9, because these are the values that appears most often in the set b of data

$$GM = \sqrt[m]{y_1 y_2 y_3 \dots y_m} = \sqrt[8]{9 \cdot 3 \cdot 8 \cdot 8 \cdot 9 \cdot 8 \cdot 9 \cdot 18} = 8.2$$ $$HM = \frac{m}{\sum_{i=1}^{m} \frac{1}{y_i}} = \frac{8}{\frac{1}{9} + \frac{1}{3} + \frac{1}{8} + \frac{1}{8} + \frac{1}{9} + \frac{1}{8} + \frac{1}{18} + \frac{1}{9}} = 7.3$$

We have an outlier in this data (18). So, the mean is the poor measure of central tendency for this data.

ii. For $a = \{12,6,7,3,15,10,18,5\}$:

$$range = x_{max} - x_{min} = 18 - 3 = 15$$

Mean deviation

$$MD = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = \frac{1}{8} (|3 - 9.5| + |5 - 9.5| + |6 - 9.5| + |7 - 9.5| + |10 - 9.5| + |12 - 9.5| + |15 - 9.5| + |18 - 9.5|) = 4.25.$$

Variance

$$V = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 = \frac{1}{8} (|3 - 9.5|^2 + |5 - 9.5|^2 + |6 - 9.5|^2 + |7 - 9.5|^2 + |10 - 9.5|^2 + |12 - 9.5|^2 + |15 - 9.5|^2 + |18 - 9.5|^2) = 23.75.$$

Standard deviation

$$SD = \sqrt{V} = \sqrt{23.75} = 4.87.$$

For $b = \{9,3,8,8,9,8,9,18\}$

$$range = y_{max} - y_{min} = 18 - 3 = 15.$$

Mean deviation

$$MD = \frac{1}{m} \sum_{i=1}^{m} |y_i - \bar{y}| = \frac{1}{8} (|3 - 9| + |8 - 9| + |8 - 9| + |9 - 9| + |9 - 9| + |9 - 9| + |18 - 9|) = 2.25.$$

Variance

$$V = \frac{1}{m} \sum_{i=1}^{m} (y_i - \bar{y})^2 =$$ $$= \frac{1}{8} (|3 - 9|^2 + |8 - 9|^2 + |8 - 9|^2 + |8 - 9|^2 + |9 - 9|^2 + |9 - 9|^2 + |9 - 9|^2 + |18 - 9|^2) = 15.$$

Standard deviation

$$SD = \sqrt{15} = 3.87.$$

The range is a poor measure of dispersion, because we have an outlier in this data (18).

iii. For $a = \{12,6,7,3,15,10,18,5\}$:

$$\begin{array}{l} \text{skewness} = \frac{1}{V^{3/2}} \cdot \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^3 = \frac{1}{23.75^2} \\ \cdot \frac{1}{8} \left((3 - 9.5)^3 + (5 - 9.5)^3 + (6 - 9.5)^3 + (7 - 9.5)^3 + (10 - 9.5)^3 + (12 - 9.5)^3 \right. \\ \left. + (15 - 9.5)^3 + (18 - 9.5)^3\right) = 0.40 \\ \end{array}$$ $$\begin{array}{l} \text{kurtosis} = \frac{1}{V^2} \cdot \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^4 = \frac{1}{23.75^2} \\ \cdot \frac{1}{8} \left(|3 - 9.5|^4 + |5 - 9.5|^4 + |6 - 9.5|^4 + |7 - 9.5|^4 + |10 - 9.5|^4 + |12 - 9.5|^4 \right. \\ \left. + |15 - 9.5|^4 + |18 - 9.5|^4\right) - 3 = -1.10 \\ \end{array}$$

For this data set, the skewness is 0.40 and the kurtosis is -1.10, which indicates small skewness and kurtosis.

For $b = \{9,3,8,8,9,8,9,18\}$:

$$\begin{array}{l} \text{skewness} = \frac{1}{V^{3/2}} \cdot \frac{1}{m} \sum_{i=1}^{m} (y_i - \bar{y})^3 = \frac{1}{15^2} \\ \cdot \frac{1}{8} \left((3 - 9)^3 + (8 - 9)^3 + (8 - 9)^3 + (8 - 9)^3 + (9 - 9)^3 + (9 - 9)^3 + (9 - 9)^3 \right. \\ \left. + (18 - 9)^3\right) = 1.10 \\ \end{array}$$ $$\begin{array}{l} \text{kurtosis} = \frac{1}{V^2} \cdot \frac{1}{m} \sum_{i=1}^{m} (y_i - \bar{y})^4 = \frac{1}{15^2} \\ \cdot \frac{1}{8} \left((3 - 9)^4 + (8 - 9)^4 + (8 - 9)^4 + (8 - 9)^4 + (9 - 9)^4 + (9 - 9)^4 + (9 - 9)^4 \right. \\ \left. + (18 - 9)^4\right) - 3 = 1.36 \\ \end{array}$$

For this data set, the skewness is 1.10 and the kurtosis is 1.36, which indicates moderate skewness and kurtosis.

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