The problem provides a table of annual revenues of a group of companies in millions of CFA francs. The table consists of revenue intervals and their corresponding frequencies. The problem asks to: 1. Determine the median, the mean, and the range of the data. Interpret the results.

1. Problem Description

The problem provides a table of annual revenues of a group of companies in millions of CFA francs. The table consists of revenue intervals and their corresponding frequencies. The problem asks to:

1. Determine the median, the mean, and the range of the data. Interpret the results.

2. Calculate the variance and standard deviation, cumulative decreasing frequencies, and the median graphically and algebraically.

2. Solution Steps

Let's denote the revenue intervals by $I_i$ and the corresponding frequencies by $n_i$.

The intervals are: [18, 30), [30, 36), [36, 42), [42, 54), [54, 60), [60, 66).

The frequencies are: 12, 90, 70, 46, 50,

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2. The total number of companies is $N = 12 + 90 + 70 + 46 + 50 + 82 = 350$.

1. Median, Mean, and Range:

a) Mean:

First, we calculate the midpoint of each interval:

$x_1 = (18+30)/2 = 24$

$x_2 = (30+36)/2 = 33$

$x_3 = (36+42)/2 = 39$

$x_4 = (42+54)/2 = 48$

$x_5 = (54+60)/2 = 57$

$x_6 = (60+66)/2 = 63$

The mean is calculated as:

$\bar{x} = \frac{\sum_{i=1}^{6} n_i x_i}{N} = \frac{12(24) + 90(33) + 70(39) + 46(48) + 50(57) + 82(63)}{350}$

$\bar{x} = \frac{288 + 2970 + 2730 + 2208 + 2850 + 5166}{350} = \frac{16212}{350} \approx 46.32$

b) Median:

The median is the value that separates the higher half from the lower half of the data. Since $N = 350$, the median is the average of the 175th and 176th values.

Cumulative frequencies:

12, 102, 172, 218, 268, 350

The 175th and 176th values fall within the interval [42, 54). To find the median, we use the formula for grouped data:

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

where:

$L$ is the lower boundary of the median interval (42)

$N$ is the total number of observations (350)

$CF$ is the cumulative frequency of the interval before the median interval (172)

$f$ is the frequency of the median interval (46)

$w$ is the width of the median interval (12)

$Median = 42 + \frac{175 - 172}{46} * 12 = 42 + \frac{3}{46} * 12 = 42 + \frac{36}{46} \approx 42 + 0.7826 \approx 42.78$

c) Range:

The range is the difference between the highest and lowest values. In this case, it's the difference between the upper limit of the last interval and the lower limit of the first interval:

Range = 66 - 18 = 48

2. Variance, Standard Deviation, Decreasing Cumulative Frequencies, and Graphical/Algebraic Median

a) Variance:

$Variance = \frac{\sum_{i=1}^{6} n_i (x_i - \bar{x})^2}{N}$

First calculate $(x_i - \bar{x})^2$ for each interval midpoint:

$(24-46.32)^2 = 507.3724$

$(33-46.32)^2 = 177.4224$

$(39-46.32)^2 = 53.5824$

$(48-46.32)^2 = 2.8224$

$(57-46.32)^2 = 114.0424$

$(63-46.32)^2 = 278.2924$

$Variance = \frac{12(507.3724) + 90(177.4224) + 70(53.5824) + 46(2.8224) + 50(114.0424) + 82(278.2924)}{350}$

$Variance = \frac{6088.4688 + 15968.016 + 3750.768 + 129.8304 + 5702.12 + 22819.9768}{350} = \frac{54459.180}{350} \approx 155.60$

b) Standard Deviation:

Standard Deviation = $\sqrt{Variance} = \sqrt{155.60} \approx 12.47$

c) Decreasing Cumulative Frequencies:

Starting from the highest interval and adding up the frequencies:

66 or less: 350

60 or less: 350 - 82 = 268

54 or less: 268 - 50 = 218

42 or less: 218 - 46 = 172

36 or less: 172 - 70 = 102

30 or less: 102 - 90 = 12

d) Median Graphically and Algebraically:

The algebraic calculation of the median was already done previously. The graphical representation would involve plotting the cumulative frequency curve and finding the value corresponding to N/2 = 175 on the x-axis.

3. Final Answer

1. Median: approximately 42.78 million CFA francs, Mean: approximately 46.32 million CFA francs, Range: 48 million CFA francs

2. Variance: approximately 155.60, Standard Deviation: approximately 12.47, Decreasing Cumulative Frequencies: 350, 268, 218, 172, 102, 12