Understanding Frequency Distribution Measures

This problem involves calculating the median of a moderately symmetrical frequency distribution using given statistical measures: the mean, the coefficient of variation, and Karl Pearson's coefficient of skewness.

Let's break down the given information:

• Mean ($\bar{x}$) = 50
• Coefficient of Variation (CV) = 35% = 0.35
• Karl Pearson's Coefficient of Skewness ($S_k$) = -0.25

We need to find the value of the Median.

Using the Coefficient of Variation (CV)

The coefficient of variation relates the standard deviation to the mean. The formula for CV is:

$$ \text{CV} = \frac{\text{Standard Deviation} (\sigma)}{\text{Mean} (\bar{x})} \times 100 $$

Given CV = 35% and Mean = 50, we can find the Standard Deviation ($\sigma$).

$$ 0.35 = \frac{\sigma}{50} $$

Solving for $\sigma$:

$$ \sigma = 50 \times 0.35 $$

$$ \sigma = 17.5 $$

The standard deviation of the distribution is 17.5.

Using Karl Pearson's Coefficient of Skewness ($S_k$)

Karl Pearson's coefficient of skewness relates the difference between the mean and the mode to the standard deviation. The formula is:

$$ S_k = \frac{\text{Mean} - \text{Mode}}{\text{Standard Deviation}} $$

Given $S_k = -0.25$, Mean = 50, and $\sigma = 17.5$, we can find the Mode.

$$ -0.25 = \frac{50 - \text{Mode}}{17.5} $$

Multiply both sides by 17.5:

$$ -0.25 \times 17.5 = 50 - \text{Mode} $$

$$ -4.375 = 50 - \text{Mode} $$

Rearrange the equation to solve for Mode:

$$ \text{Mode} = 50 + 4.375 $$

$$ \text{Mode} = 54.375 $$

The mode of the distribution is 54.375.

Finding the Median using Empirical Relation

For a moderately asymmetrical frequency distribution, there is an empirical relationship between the Mean, Median, and Mode:

$$ \text{Mean} - \text{Mode} \approx 3 \times (\text{Mean} - \text{Median}) $$

We have the Mean = 50 and Mode = 54.375. We can substitute these values into the formula to find the Median.

$$ 50 - 54.375 = 3 \times (50 - \text{Median}) $$

$$ -4.375 = 3 \times (50 - \text{Median}) $$

Divide both sides by 3:

$$ \frac{-4.375}{3} = 50 - \text{Median} $$

$$ -1.4583... \approx 50 - \text{Median} $$

Rearrange to solve for Median:

$$ \text{Median} \approx 50 + 1.4583... $$

$$ \text{Median} \approx 51.4583... $$

Comparing with the Options

Now, let's evaluate the given options to see which one matches our calculated median value ($\approx 51.4583...$).

Comparing the decimal values, option 3, \(\dfrac{1235}{24}\), is approximately 51.4583..., which matches our calculated value for the median.

Conclusion

Using the given mean, coefficient of variation, and Karl Pearson's coefficient of skewness, and applying the empirical relationship for a moderately asymmetrical distribution, we found the median to be approximately 51.4583..., which corresponds to the fraction \(\dfrac{1235}{24}\).

Revision Table: Frequency Distribution Measures

Additional Information: Skewness and Distribution Shapes

Skewness indicates the degree of asymmetry in a frequency distribution. Understanding skewness helps in interpreting the relationship between the mean, median, and mode.

• Symmetrical Distribution: In a perfectly symmetrical distribution (like a normal distribution), the Mean, Median, and Mode are all equal. The skewness is 0.
• Positively Skewed Distribution (Right Skewed): The tail of the distribution is longer on the right side. Mean > Median > Mode. Karl Pearson's coefficient of skewness is positive.
• Negatively Skewed Distribution (Left Skewed): The tail of the distribution is longer on the left side. Mode > Median > Mean. Karl Pearson's coefficient of skewness is negative.

In this problem, the Karl Pearson's coefficient of skewness is -0.25, which is negative. This indicates a negatively skewed distribution, meaning the Mode is generally greater than the Median, and the Median is generally greater than the Mean (Mode > Median > Mean). Our calculated values align with this: Mode (54.375) > Median (51.4583...) > Mean (50).

The empirical formula used, \(\text{Mean} - \text{Mode} \approx 3(\text{Mean} - \text{Median})\), is a good approximation for moderately skewed distributions, which is stated in the problem.