1. **Problem:** Calculate the arithmetic mean (population mean, sample mean, median) of the Education Level data. 2. **Data:** Education Level values from the dataset: $[1,2,1,1,1,1,1,1,3,1,1,1,1,1,3,2,1,2,3,2,2,1,2,2,2]$ 3. **Arithmetic Mean (Population and Sample Mean):** The formula for the mean is: $$\text{Mean} = \frac{\sum x_i}{n}$$ where $x_i$ are the data points and $n$ is the number of data points. 4. **Calculate the sum:** $$\sum x_i = 1+2+1+1+1+1+1+1+3+1+1+1+1+1+3+2+1+2+3+2+2+1+2+2+2 = 41$$ 5. **Count the number of data points:** $$n = 25$$ 6. **Calculate the mean:** $$\text{Mean} = \frac{41}{25} = 1.64$$ 7. **Median:** Sort the data: $$[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3]$$ The middle value (13th in sorted list) is $1$. **Median = 1** --- 8. **Weighted Mean:** Weights are frequencies of each education level: - Bachelor Degree (1): count how many 1's - Higher Degree (2): count how many 2's - High School (3): count how many 3's Count: - 1 appears 15 times - 2 appears 8 times - 3 appears 3 times Weighted mean formula: $$\text{Weighted Mean} = \frac{\sum w_i x_i}{\sum w_i}$$ where $w_i$ is the frequency. Calculate: $$\frac{15\times1 + 8\times2 + 3\times3}{15+8+3} = \frac{15 + 16 + 9}{26} = \frac{40}{26} \approx 1.5385$$ --- 9. **Geometric Mean:** Formula: $$\text{Geometric Mean} = \sqrt[n]{\prod_{i=1}^n x_i}$$ Calculate product: $$1^{15} \times 2^{8} \times 3^{3} = 1 \times 256 \times 27 = 6912$$ Number of data points $n=26$ (corrected count from weighted mean denominator) Calculate geometric mean: $$\sqrt[26]{6912} = e^{\frac{\ln(6912)}{26}} \approx e^{\frac{8.840}{26}} = e^{0.34} \approx 1.405$$ --- 10. **Measure of Dispersion:** - Range = max - min = $3 - 1 = 2$ - Sample Variance formula: $$s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2$$ Calculate squared deviations: For $x=1$: $(1-1.64)^2 = 0.4096$, count 15 times For $x=2$: $(2-1.64)^2 = 0.1296$, count 8 times For $x=3$: $(3-1.64)^2 = 1.8496$, count 3 times Sum: $$15 \times 0.4096 + 8 \times 0.1296 + 3 \times 1.8496 = 6.144 + 1.0368 + 5.5488 = 12.7296$$ Sample variance: $$s^2 = \frac{12.7296}{25} = 0.5092$$ Sample standard deviation: $$s = \sqrt{0.5092} \approx 0.7136$$ --- 11. **Coefficient of Skewness:** Formula (Pearson's first coefficient): $$\text{Skewness} = \frac{3(\bar{x} - \text{Median})}{s}$$ Calculate: $$\frac{3(1.64 - 1)}{0.7136} = \frac{3 \times 0.64}{0.7136} = \frac{1.92}{0.7136} \approx 2.69$$ Positive skewness indicates right-skewed distribution. --- 12. **Chebyshev's Theorem:** For any $k > 1$, at least $\left(1 - \frac{1}{k^2}\right)100\%$ of data lies within $k$ standard deviations of the mean. Example: For $k=2$, at least $1 - \frac{1}{4} = 0.75$ or 75% of data lies within $2s$ of mean. Calculate interval: $$[\bar{x} - 2s, \bar{x} + 2s] = [1.64 - 2(0.7136), 1.64 + 2(0.7136)] = [0.213, 3.067]$$ All data points (1,2,3) lie within this interval. --- 13. **Empirical Rule:** For approximately normal data: - 68% within $1s$ - 95% within $2s$ - 99.7% within $3s$ Calculate $1s$ interval: $$[1.64 - 0.7136, 1.64 + 0.7136] = [0.9264, 2.3536]$$ Count data points within this range: - Values 1 and 2 are within range, 3 is outside. Percentage: $$\frac{15 + 8}{26} = \frac{23}{26} \approx 88.5\%$$ This suggests data is not perfectly normal but concentrated around mean. --- **Final answers:** - Population/Sample Mean: $1.64$ - Median: $1$ - Weighted Mean: $1.5385$ - Geometric Mean: $1.405$ - Range: $2$ - Sample Variance: $0.5092$ - Sample Standard Deviation: $0.7136$ - Coefficient of Skewness: $2.69$ - Chebyshev's theorem interval for $k=2$: $[0.213, 3.067]$ - Empirical rule $1s$ interval: $[0.9264, 2.3536]$ with approx 88.5% data inside