Question

A panel is conducting an interview on six candidates of different heights. If they are to put them in line, in how many ways can they arrange them in line such that no three consecutive candidates are in increasing order of height from front to back?

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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