BCO127 Applied Management Statistics Task Brief & Rubrics

BCO127 Applied Management Statistics Task brief & rubrics Task

In a sample of 30 individuals from a city, their income data are as follows: 1 person with no income, 2 earning 500 €, 3 earning 1000 €, 5 earning 1500 €, 8 earning 2000 €, 5 earning 2500 €, 3 earning 3000 €, 2 earning 3500 €, and 1 earning 4000 €. Based on this data, perform the following statistical analyses:

• Calculate the arithmetic mean and the geometric mean of the income data.
• Determine the median and the mode of the income distribution.
• Calculate the deviation of each income value from the mean and find the total sum of these deviations.
• Compute the variance and the standard deviation of the data.
• Draw a column chart to visualize the distribution based on the frequencies.
• Assuming the data is representative of a normally distributed population, find the probability that a randomly selected individual earns between 1062 € and 2938 €.
• Similarly, find the probability that a randomly chosen individual has a monthly income less than 1062 €.

Complete this task with a report of approximately 1000 words, including all steps, calculations, and comments, formatted in Arial 12 pt font, justified alignment, and Harvard citation style for references. Demonstrate understanding of key statistical concepts, formulas, and techniques; clearly explain your reasoning; and present your results accurately and professionally.

Paper For Above instruction

The analysis of income distribution within a population provides vital insights into economic well-being, disparity, and the overall shape of income variation. Using sample data from 30 individuals residing in a city, this report conducts a comprehensive statistical assessment, including measures of central tendency, variability, distribution visualization, and probability calculations assuming normality. Each step is carefully detailed to ensure clarity and transparency in the methodological approach.

Introduction

Understanding income distribution is a fundamental aspect of economic statistics, informing policy decisions and sociological theories. This report explores various descriptive and inferential statistics based on a provided sample dataset. The goal is to interpret the data thoroughly, applying fundamental formulas, and understanding the implications of the findings in a broader context. Central measures such as the mean, median, mode, and measures of dispersion like variance and standard deviation are computed, alongside probability estimations under the assumption of normality.

Data Summary

The data set consists of income values categorized into frequency counts provided in the sample. Specifically, the class frequencies are:

• No income: 1 person
• €500 income: 2 persons
• €1000 income: 3 persons
• €1500 income: 5 persons
• €2000 income: 8 persons
• €2500 income: 5 persons
• €3000 income: 3 persons
• €3500 income: 2 persons
• €4000 income: 1 person

These frequencies sum to 30, representing the entire sample size.

Calculation of Measures of Central Tendency

Arithmetic Mean

The arithmetic mean (average) income is calculated as the sum of all incomes weighted by their frequencies divided by the total number of observations (30). Mathematically:

Mean = (Σ f_i * x_i ) / N

Where f_i are the frequencies and x_i are the income values.

Calculating:

• 0 € * 1 = 0
• 500 € * 2 = 1000 €
• 1000 € * 3 = 3000 €
• 1500 € * 5 = 7500 €
• 2000 € * 8 = 16,000 €
• 2500 € * 5 = 12,500 €
• 3000 € * 3 = 9,000 €
• 3500 € * 2 = 7,000 €
• 4000 € * 1 = 4,000 €

Total sum: 0 + 1000 + 3000 + 7500 + 16,000 + 12,500 + 9,000 + 7,000 + 4,000 = 62,000 €

Mean income = 62,000 € / 30 ≈ 2066.67 €

Geometric Mean

The geometric mean (GM) accounts for multiplicative relationships and is suitable for skewed data like income distributions. It is computed as:

GM = (Π x_i ^ f_i)^{1 / N}

which is often easier to compute using logs:

GM = exp[(1 / N) Σ f_i ln(x_i)]

Calculations:

• ln(0) is undefined; income of zero is considered as zero, but for geometric mean, zero incomes present a special case. Typically, geometric mean is undefined with zero values; in this case, to avoid mathematical issues, income zero can be treated as a small positive number (e.g., 1 €). Alternatively, one may exclude zero income from geometric mean calculations or note that GM cannot be computed with zeros, considering the relevance of the context.

Assuming we treat zero as 1 € (least positive amount), then ln(1) = 0. Recalculating accordingly or choosing to omit zero income from GM calculation is standard practice in income analysis, as geometric mean excludes zeros. Here, for simplicity, we omit the zero income value; thus, the sum considers only positive incomes:

• For incomes > 0 €, the sum of f_i * ln(x_i):

(2 ln(500)) + (3 ln(1000)) + (5 ln(1500)) + (8 ln(2000)) + (5 ln(2500)) + (3 ln(3000)) + (2 ln(3500)) + (1 ln(4000))

Calculating logs (approximated):

• ln(500) ≈ 6.2146
• ln(1000) ≈ 6.9088
• ln(1500) ≈ 7.3132
• ln(2000) ≈ 7.6014
• ln(2500) ≈ 7.8234
• ln(3000) ≈ 8.0064
• ln(3500) ≈ 8.1605
• ln(4000) ≈ 8.2940

Sum of weighted logs: (26.2146) + (36.9088) + (57.3132) + (87.6014) + (57.8234) + (38.0064) + (28.1605) + (18.2940) ≈ 12.4292 + 20.7264 + 36.566 + 60.8112 + 39.117 + 24.0192 + 16.321 + 8.294 ≈ 218.265

Therefore, GM = exp(218.265 / 30) ≈ exp(7.2755) ≈ 1,444.25 €

Median and Mode

Median

The median is the middle value when data are ordered. With frequencies, locate the position of the median in the cumulative distribution:

• Cumulative frequency up to income €500: 2
• Up to €1000: 5
• Up to €1500: 10
• Up to €2000: 18
• Up to €2500: 23
• Up to €3000: 26
• Up to €3500: 28
• Up to €4000: 29

The median position is at (N+1)/2 = 15.5. The 15th value lies within the cumulative frequency of 18, corresponding to €2000. Since 15.5 is between 10 and 18, the median income is approximately €2000.

Mode

The mode is the income value with the highest frequency, which is 2000 € with 8 occurrences.

Deviations and Total Sum of Deviations

Deviations are calculated for each income value as:

deviation = income - mean ≈ income - 2066.67 €

• No income: 0 - 2066.67 ≈ -2066.67
• €500: 500 - 2066.67 ≈ -1566.67
• €1000: 1000 - 2066.67 ≈ -1066.67
• €1500: 1500 - 2066.67 ≈ -566.67
• €2000: 2000 - 2066.67 ≈ -66.67
• €2500: 2500 - 2066.67 ≈ 433.33
• €3000: 3000 - 2066.67 ≈ 933.33
• €3500: 3500 - 2066.67 ≈ 1433.33
• €4000: 4000 - 2066.67 ≈ 1933.33

Each deviation is multiplied by the number of persons with that income, and the sum of all deviations (Σ deviations) should theoretically be close to zero due to the properties of the mean, but slight discrepancies may exist due to rounding:

Total deviation sum ≈ (−2066.671) + (−1566.672) + (−1066.673) + (−566.675) + (−66.678) + (433.335) + (933.333) + (1433.332) + (1933.33*1) ≈ -2066.67 - 3133.34 - 3199 - 2833.35 - 533.36 + 2166.65 + 2800 - 2866.66 + 1933.33 ≈ 0 (as expected, minor rounding errors included).

Variability: Variance and Standard Deviation

Variance

The variance measures dispersion around the mean and is calculated as:

Variance = (Σ f_i * (x_i - mean)²) / (N - 1)

Calculations:

• (0 - 2066.67)² * 1 ≈ 4,273,605
• (500 - 2066.67)² 2 ≈ (−1566.67)² 2 ≈ 4,455,555
• (1000 - 2066.67)² 3 ≈ (−1066.67)² 3 ≈ 3,425,560
• (1500 - 2066.67)² 5 ≈ (−566.67)² 5 ≈ 1,607,650
• (2000 - 2066.67)² 8 ≈ (−66.67)² 8 ≈ 35,558
• (2500 - 2066.67)² 5 ≈ 433.33² 5 ≈ 938,889
• (3000 - 2066.67)² 3 ≈ 933.33² 3 ≈ 2,616,660
• (3500 - 2066.67)² 2 ≈ 1,433.33² 2 ≈ 4,112,589
• (4000 - 2066.67)² * 1 ≈ 1,933.33² ≈ 3,736,607

Sum of squared deviations: approximately 22,557,177

Variance ≈ 22,557,177 / (30 - 1) ≈ 776,454

Standard Deviation

Standard deviation is the square root of variance:

SD ≈ √776,454 ≈ 880.00 €

Distribution Visualization

A column chart illustrating the frequency of each income category visually depicts the income distribution. The highest frequency occurs at €2000, indicating a concentration of earners around this amount, with a decreasing trend towards the extremes of no income and €4000. This pattern suggests a skewed distribution with a peak near the median income, consistent with typical income data profiles.

Probability Under Normal Distribution

Assuming the income data follow a normal distribution with mean ≈ 2066.67 € and SD ≈ 880 €, the probabilities are computed using standard z-scores:

Probability between €1062 and €2938

z = (x - mean) / SD

• Z₁ = (1062 - 2066.67) / 880 ≈ -1.13
• Z₂ = (2938 - 2066.67) / 880 ≈ 1.02

Using standard normal tables or calculators:

P(1062

Approximately 71.7% probability that a randomly selected individual earns between €1062 and €2938.

Probability income

Z = -1.13 (same as above)

P(X

This indicates a roughly 13% chance of selecting an individual with income below €1062.

Conclusion

This analysis demonstrates how descriptive statistics can illuminate the characteristics of income distribution, revealing central tendency, variability, and dispersion. The use of the normal distribution model permits probability estimations, providing insights into income ranges, though caution is advised given the skewness typical of income data. Overall, the methods applied demonstrate core statistical concepts essential for economic and management analyses, reinforcing the importance of accurate calculation, interpretation, and visualization in decision-making processes.

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