1. **State the problem:** Calculate Karl Pearson's Coefficient of Skewness for the given frequency distribution. 2. **Given data:** Class intervals: 30-35, 35-40, 40-45, 45-50, 50-55, 55-60 Frequencies: 5, 10, 30, 35, 15, 5 3. **Find the midpoints (x) of each class:** $$30-35 \rightarrow 32.5$$ $$35-40 \rightarrow 37.5$$ $$40-45 \rightarrow 42.5$$ $$45-50 \rightarrow 47.5$$ $$50-55 \rightarrow 52.5$$ $$55-60 \rightarrow 57.5$$ 4. **Calculate the mean ($\bar{x}$):** $$\bar{x} = \frac{\sum f x}{\sum f}$$ Calculate $\sum f x$: $$5 \times 32.5 = 162.5$$ $$10 \times 37.5 = 375$$ $$30 \times 42.5 = 1275$$ $$35 \times 47.5 = 1662.5$$ $$15 \times 52.5 = 787.5$$ $$5 \times 57.5 = 287.5$$ Sum of frequencies $\sum f = 5 + 10 + 30 + 35 + 15 + 5 = 100$ Sum of $f x = 162.5 + 375 + 1275 + 1662.5 + 787.5 + 287.5 = 4350$ Mean: $$\bar{x} = \frac{4350}{100} = 43.5$$ 5. **Find the median class:** Total frequency = 100 Median position = $\frac{100}{2} = 50$ Cumulative frequencies: - 30-35: 5 - 35-40: 15 - 40-45: 45 - 45-50: 80 Median class is 45-50 because cumulative frequency just before it is 45 and after is 80. 6. **Calculate median ($M$):** Using formula: $$M = L + \left(\frac{\frac{N}{2} - F}{f_m}\right) \times h$$ Where: - $L = 45$ (lower boundary of median class) - $N = 100$ - $F = 45$ (cumulative frequency before median class) - $f_m = 35$ (frequency of median class) - $h = 5$ (class width) Calculate: $$M = 45 + \left(\frac{50 - 45}{35}\right) \times 5 = 45 + \left(\frac{5}{35}\right) \times 5 = 45 + \frac{25}{35} = 45 + 0.714 = 45.714$$ 7. **Find the mode ($Mo$):** Modal class is the class with highest frequency = 45-50 with frequency 35. Using formula: $$Mo = L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$ Where: - $L = 45$ - $f_1 = 35$ (frequency of modal class) - $f_0 = 30$ (frequency before modal class) - $f_2 = 15$ (frequency after modal class) - $h = 5$ Calculate: $$Mo = 45 + \left(\frac{35 - 30}{2 \times 35 - 30 - 15}\right) \times 5 = 45 + \left(\frac{5}{70 - 45}\right) \times 5 = 45 + \left(\frac{5}{25}\right) \times 5 = 45 + 1 = 46$$ 8. **Calculate Karl Pearson's Coefficient of Skewness ($Sk$):** $$Sk = \frac{\bar{x} - Mo}{\sigma}$$ But since standard deviation ($\sigma$) is not given, Karl Pearson's formula often uses: $$Sk = \frac{\bar{x} - Mo}{\text{standard deviation}}$$ Alternatively, the formula using mean, median, and standard deviation is: $$Sk = \frac{3(\bar{x} - M)}{\sigma}$$ Since $\sigma$ is not given, we use the formula: $$Sk = \frac{\bar{x} - Mo}{\sigma}$$ We need to calculate standard deviation. 9. **Calculate variance and standard deviation:** Calculate $\sum f x^2$: $$5 \times 32.5^2 = 5 \times 1056.25 = 5281.25$$ $$10 \times 37.5^2 = 10 \times 1406.25 = 14062.5$$ $$30 \times 42.5^2 = 30 \times 1806.25 = 54187.5$$ $$35 \times 47.5^2 = 35 \times 2256.25 = 78968.75$$ $$15 \times 52.5^2 = 15 \times 2756.25 = 41343.75$$ $$5 \times 57.5^2 = 5 \times 3306.25 = 16531.25$$ Sum: $$\sum f x^2 = 5281.25 + 14062.5 + 54187.5 + 78968.75 + 41343.75 + 16531.25 = 210375$$ Variance: $$\sigma^2 = \frac{\sum f x^2}{\sum f} - \bar{x}^2 = \frac{210375}{100} - 43.5^2 = 2103.75 - 1892.25 = 211.5$$ Standard deviation: $$\sigma = \sqrt{211.5} \approx 14.54$$ 10. **Calculate Karl Pearson's Coefficient of Skewness:** $$Sk = \frac{\bar{x} - Mo}{\sigma} = \frac{43.5 - 46}{14.54} = \frac{-2.5}{14.54} \approx -0.172$$ **Final answer:** Karl Pearson's Coefficient of Skewness is approximately **$-0.172$**, indicating a slight negative skewness.