The problem presents a table showing the annual profit (in millions of FCFA) of 170 Senegalese companies, grouped into classes. It asks several questions related to statistical analysis: 1. Identify the population, statistical unit, observed variable, and its nature.

1. Problem Description

The problem presents a table showing the annual profit (in millions of FCFA) of 170 Senegalese companies, grouped into classes. It asks several questions related to statistical analysis:

1. Identify the population, statistical unit, observed variable, and its nature.

2. Construct the histogram and frequency polygon.

3. Construct the increasing and decreasing cumulative frequency curves.

4. Calculate the arithmetic mean, standard deviation, and coefficient of variation.

5. Determine the modal class, median, and mean deviation. Deduce the concentration gap.

6. Calculate the Gini concentration index and draw the curve.

2. Solution Steps

1. Population, Statistical Unit, Variable, and Nature:

* Population: The 170 Senegalese companies.

* Statistical unit: Each Senegalese company.

* Observed variable: Annual profit (in millions of FCFA).

* Nature: Quantitative continuous (since profit can take on a range of values within each interval).

2. Histogram and Frequency Polygon:

To construct the histogram and frequency polygon, we need to use the class intervals and frequencies given. The histogram will have bars representing each class, with the height of each bar proportional to the frequency of that class. The frequency polygon is formed by connecting the midpoints of the top of each bar in the histogram.

Class Interval | Frequency

------- | --------

[10, 30[ | 55

[30, 50[ | 45

[50, 60[ | 30

[60, 75[ | 22

[75, 90[ | 18

Midpoints of the classes are: 20, 40, 55, 67.5, 82.5

3. Increasing and Decreasing Cumulative Frequency Curves:

To construct the cumulative frequency curves, we need to calculate the cumulative frequencies.

Class Interval | Frequency | Cumulative Frequency (Increasing) | Cumulative Frequency (Decreasing)

------- | -------- | -------- | --------

[10, 30[ | 55 | 55 | 170

[30, 50[ | 45 | 100 | 115

[50, 60[ | 30 | 130 | 70

[60, 75[ | 22 | 152 | 40

[75, 90[ | 18 | 170 | 18

The increasing cumulative frequency curve starts at 0 and increases with each class. The decreasing cumulative frequency curve starts at the total frequency (170) and decreases with each class.

4. Arithmetic Mean, Standard Deviation, and Coefficient of Variation:

First, calculate the midpoint of each class:

$x_1 = (10+30)/2 = 20$

$x_2 = (30+50)/2 = 40$

$x_3 = (50+60)/2 = 55$

$x_4 = (60+75)/2 = 67.5$

$x_5 = (75+90)/2 = 82.5$

The arithmetic mean ($\bar{x}$) is calculated as:

$\bar{x} = \frac{\sum{f_i * x_i}}{\sum{f_i}}$

$\bar{x} = (55*20 + 45*40 + 30*55 + 22*67.5 + 18*82.5) / 170$

$\bar{x} = (1100 + 1800 + 1650 + 1485 + 1485) / 170 = 7520 / 170 \approx 44.24$

To calculate the standard deviation ($s$), we first calculate the variance ($s^2$):

$s^2 = \frac{\sum{f_i * (x_i - \bar{x})^2}}{\sum{f_i}}$

$s^2 = [55*(20-44.24)^2 + 45*(40-44.24)^2 + 30*(55-44.24)^2 + 22*(67.5-44.24)^2 + 18*(82.5-44.24)^2]/170$

$s^2 = [55*587.5876 + 45*17.9776 + 30*115.7776 + 22*541.8276 + 18*1463.4276]/170$

$s^2 = [32317.318 +

8

0

9. 0 + 3473.328 + 11920.2072 + 26341.6968]/170$

$s^2 = 74861.55 / 170 \approx 440.36$

The standard deviation is:

$s = \sqrt{s^2} = \sqrt{440.36} \approx 20.98$

The coefficient of variation ($CV$) is:

$CV = \frac{s}{\bar{x}} = \frac{20.98}{44.24} \approx 0.47$ or 47%

5. Modal Class, Median, and Mean Deviation:

* Modal class: The class with the highest frequency, which is [10, 30[ (frequency = 55).

* Median: The median is the value that separates the higher half from the lower half of the data set. Since the total frequency is 170, the median lies within the class containing the 85th and 86th values. The cumulative frequency reaches 55 at the end of the first class [10, 30[. It reaches 100 at the end of the second class [30, 50[. Therefore, the median lies in the [30, 50[ class. Using interpolation:

$Median = L + (\frac{N/2 - CF_{below}}{f_{median}}) * w$

where:

$L$ is the lower limit of the median class (30)

$N$ is the total frequency (170)

$CF_{below}$ is the cumulative frequency of the class before the median class (55)

$f_{median}$ is the frequency of the median class (45)

$w$ is the class width (20)

$Median = 30 + (\frac{170/2 - 55}{45}) * 20 = 30 + (\frac{85-55}{45}) * 20 = 30 + (\frac{30}{45}) * 20 = 30 + (2/3)*20 = 30 + 13.33 = 43.33$

* Mean deviation: Since the data is grouped, the mean deviation isn't directly calculable without access to the original data. We would need the values of each company's profit to compute the mean deviation correctly. Therefore, we can only estimate the mean deviation using grouped data.

6. Gini Concentration Index and Curve:

The Gini index cannot be accurately computed from the given grouped data without making assumptions about the distribution within each class interval. Usually, for such calculations, one would either need the original data or assume uniform distribution within each class. Because the Gini index calculation requires the cumulative proportion of companies and their cumulative proportion of profit, and because we do not have the overall total profit we can not accurately calculate this index.

3. Final Answer

1. Population: 170 Senegalese companies. Statistical unit: Each Senegalese company. Observed variable: Annual profit (in millions of FCFA). Nature: Quantitative continuous.

2. Histogram and frequency polygon: See explanation in solution steps.

3. Increasing and decreasing cumulative frequency curves: See explanation in solution steps.

4. Arithmetic mean: approximately 44.24 million FCFA. Standard deviation: approximately 20.98 million FCFA. Coefficient of variation: approximately 47%.

5. Modal class: [10, 30[. Median: approximately 43.33 million FCFA. Mean deviation can only be estimated.

6. Gini index cannot be accurately calculated with only grouped data.