The problem provides a table showing the distribution of inhabitants of a municipality based on their annual local taxes in thousands of CFA francs. The questions ask us to identify the population studied, the statistical unit, and the observed character (variable). Then, we need to construct a histogram, a frequency polygon, a cumulative increasing frequency curve, and a cumulative decreasing frequency curve. Finally, we need to determine the mode, median, quartiles Q1 and Q3, and interpret them. Also, we need to calculate the arithmetic, geometric, harmonic, and quadratic means.

1. Problem Description

The problem provides a table showing the distribution of inhabitants of a municipality based on their annual local taxes in thousands of CFA francs. The questions ask us to identify the population studied, the statistical unit, and the observed character (variable). Then, we need to construct a histogram, a frequency polygon, a cumulative increasing frequency curve, and a cumulative decreasing frequency curve. Finally, we need to determine the mode, median, quartiles Q1 and Q3, and interpret them. Also, we need to calculate the arithmetic, geometric, harmonic, and quadratic means.

2. Solution Steps

1. Population studied: The inhabitants of a municipality.

Statistical unit: Each inhabitant of the municipality.

Observed character: The amount of annual local taxes in thousands of CFA francs.

Nature of the character: Quantitative continuous (since the amount of tax can take any value within a range).

2. To construct the histogram, frequency polygon, and cumulative frequency curves, we need to calculate the class widths, frequencies, cumulative frequencies, and cumulative relative frequencies. We also need to choose appropriate scales for the axes. Since actually drawing them is impossible in this text based format, a detailed textual explanation follows:

* Histogram: The x-axis represents the tax brackets (classes), and the y-axis represents the frequencies (number of inhabitants in each class). The width of each rectangle corresponds to the class width. The height corresponds to the frequency. For classes with different widths, the height of the rectangle should be adjusted so that the area is proportional to the frequency (frequency density).

* Frequency Polygon: We plot the midpoints of each class against the corresponding frequencies. Connect consecutive points with straight lines.

* Cumulative Increasing Frequency Curve (Ogive): The x-axis represents the upper limits of the tax brackets, and the y-axis represents the cumulative frequencies (the number of inhabitants with taxes less than or equal to the upper limit).

* Cumulative Decreasing Frequency Curve: The x-axis represents the lower limits of the tax brackets, and the y-axis represents the cumulative frequencies (the number of inhabitants with taxes greater than or equal to the lower limit).

3. Determining the Mode, Median, and Quartiles:

* Mode: The modal class is the class with the highest frequency. In this case, the class [12, 16[ has the highest frequency (19). Assuming a uniform distribution within the class, we can approximate the mode as the midpoint of the modal class: $(12+16)/2 = 14$.

* Median: The median is the value that divides the distribution into two equal halves. First, we find the total number of inhabitants: $N = 1 + 7 + 11 + 8 + 12 + 15 + 19 + 16 + 8 + 3 = 100$. The median is the value corresponding to the 50th inhabitant ($N/2 = 50$). We look for the class where the cumulative frequency reaches or exceeds

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0.

The cumulative frequencies are:

1, 8, 19, 27, 39, 54, 73, 89, 97,

1

0

0.

The median class is [10, 12[. We interpolate to find the median value:

$Median = L + (\frac{N/2 - CF}{f}) * w$

where $L$ is the lower limit of the median class (10), $N$ is the total number of observations (100), $CF$ is the cumulative frequency of the class before the median class (39), $f$ is the frequency of the median class (15), and $w$ is the width of the median class (2).

$Median = 10 + (\frac{50 - 39}{15}) * 2 = 10 + (\frac{11}{15}) * 2 \approx 10 + 1.47 = 11.47$

* Q1: The first quartile (Q1) is the value that divides the distribution such that 25% of the data is below it. We find the class where the cumulative frequency reaches or exceeds $N/4 = 100/4 = 25$. The class is [6, 8[.

$Q1 = L + (\frac{N/4 - CF}{f}) * w$

where $L$ is the lower limit of the Q1 class (6), $N$ is the total number of observations (100), $CF$ is the cumulative frequency of the class before the Q1 class (8), $f$ is the frequency of the Q1 class (11), and $w$ is the width of the Q1 class (2).

$Q1 = 6 + (\frac{25 - 8}{11}) * 2 = 6 + (\frac{17}{11}) * 2 \approx 6 + 3.09 = 9.09$

* Q3: The third quartile (Q3) is the value that divides the distribution such that 75% of the data is below it. We find the class where the cumulative frequency reaches or exceeds $3N/4 = 75$. The class is [12, 16[.

$Q3 = L + (\frac{3N/4 - CF}{f}) * w$

where $L$ is the lower limit of the Q3 class (12), $N$ is the total number of observations (100), $CF$ is the cumulative frequency of the class before the Q3 class (54), $f$ is the frequency of the Q3 class (19), and $w$ is the width of the Q3 class (4).

$Q3 = 12 + (\frac{75 - 54}{19}) * 4 = 12 + (\frac{21}{19}) * 4 \approx 12 + 4.42 = 16.42$

4. Calculating the means:

To calculate the different types of means, we first need to find the midpoints ($x_i$) of each class: 3, 5, 7, 8.5, 9.5, 11, 14, 18, 30,

6

0. The frequencies ($f_i$) are: 1, 7, 11, 8, 12, 15, 19, 16, 8,

3. $N = \sum f_i = 100$.

* Arithmetic Mean:

$\bar{x} = \frac{\sum x_i f_i}{N} = \frac{(3*1 + 5*7 + 7*11 + 8.5*8 + 9.5*12 + 11*15 + 14*19 + 18*16 + 30*8 + 60*3)}{100} = \frac{3 + 35 + 77 + 68 + 114 + 165 + 266 + 288 + 240 + 180}{100} = \frac{1436}{100} = 14.36$

* Geometric Mean:

$GM = (\prod x_i^{f_i})^{1/N} = (3^1 * 5^7 * 7^{11} * 8.5^8 * 9.5^{12} * 11^{15} * 14^{19} * 18^{16} * 30^8 * 60^3)^{1/100}$

Taking the logarithm:

$log(GM) = \frac{1}{100} * (1*log(3) + 7*log(5) + 11*log(7) + 8*log(8.5) + 12*log(9.5) + 15*log(11) + 19*log(14) + 16*log(18) + 8*log(30) + 3*log(60))$

$log(GM) = \frac{1}{100} * (0.477 + 4.898 + 9.332 + 7.423 + 11.158 + 15.665 + 21.527 + 20.643 + 11.842 + 4.669) = \frac{107.634}{100} = 1.07634$

$GM = 10^{1.07634} \approx 11.92$

* Harmonic Mean:

$HM = \frac{N}{\sum \frac{f_i}{x_i}} = \frac{100}{\frac{1}{3} + \frac{7}{5} + \frac{11}{7} + \frac{8}{8.5} + \frac{12}{9.5} + \frac{15}{11} + \frac{19}{14} + \frac{16}{18} + \frac{8}{30} + \frac{3}{60}} = \frac{100}{0.333 + 1.4 + 1.571 + 0.941 + 1.263 + 1.364 + 1.357 + 0.889 + 0.267 + 0.05} = \frac{100}{9.435} \approx 10.598$

* Quadratic Mean:

$QM = \sqrt{\frac{\sum x_i^2 f_i}{N}} = \sqrt{\frac{3^2*1 + 5^2*7 + 7^2*11 + 8.5^2*8 + 9.5^2*12 + 11^2*15 + 14^2*19 + 18^2*16 + 30^2*8 + 60^2*3}{100}} = \sqrt{\frac{9 + 175 + 539 + 578 + 1083 + 1815 + 3724 + 5184 + 7200 + 10800}{100}} = \sqrt{\frac{31107}{100}} = \sqrt{311.07} \approx 17.64$

3. Final Answer

1. Population studied: The inhabitants of the municipality. Statistical unit: Each inhabitant of the municipality. Observed character: The amount of annual local taxes in thousands of CFA francs. Nature of the character: Quantitative continuous.

2. See Solution Steps for a descriptive explanation.

3. Mode: 14 (thousand CFA francs). Median: 11.47 (thousand CFA francs). Q1: 9.09 (thousand CFA francs). Q3: 16.42 (thousand CFA francs).

4. Arithmetic Mean: 14.36 (thousand CFA francs). Geometric Mean: 11.92 (thousand CFA francs). Harmonic Mean: 10.60 (thousand CFA francs). Quadratic Mean: 17.64 (thousand CFA francs).