Consider the following frequency distribution: class: 10-20, 20-30, 30-40, 40-50, 50-60; Frequency:

Q: Wait, why is the median class $40-50$?

The median is given as $45$. In a grouped frequency distribution, the median class is the interval that contains the median value. Since $45$ lies between $40$ and $50$, the class $40-50$ is our median class.

Q: What is $cf$ exactly? Is it just the frequency of the class before?

Great question! In the median formula, $cf$ stands for the *cumulative* frequency of the class *preceding* the median class. Since our median class is $40-50$, we sum all frequencies before it: $\alpha + 110 + 54 = \alpha + 164$. It is a running total, not just a single frequency!

Q: Is it possible for the median to fall outside the median class?

Conceptually, no. The median is a measure of central tendency. By definition, the median class is the interval where the cumulative frequency first exceeds $\frac{N}{2}$. If your calculated median falls outside your chosen class, you have likely misidentified the median class or made an error in your cumulative frequency calculation.

Question

Consider the following frequency distribution $$\begin{array}{|c|c|c|c|c|c|} \hline class : & 10 - 20 & 20 - 30 & 30 - 40 & 40 - 50 & 50 - 60 \ \hline Frequency : & \alpha & 110 & 54 & 30 & \beta \ \hline \end{array}$$ 
If the sum of all frequencies is 584 and median is 45, then $|\alpha - \beta|$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the world of mathematics. Today, we are going to unravel a classic problem in statistics.

We are given a total frequency $N = 584$ and a median $M = 45$. Our mission is to find the absolute difference between two mysterious variables, $\alpha$ and $\beta$.

### The First Constraint: The Total Sum

The first piece of information we can use is the total frequency. If we add up all the values in the frequency column, their sum must equal $584$.

We write the equation:
$$\alpha + 110 + 54 + 30 + \beta = 584$$

Adding the known numbers, $110 + 54 + 30$, gives us $194$. Thus, the equation simplifies to:
$$\alpha + \beta + 194 = 584$$

Subtracting $194$ from both sides, we obtain:
$$\alpha + \beta = 390$$
Let us call this **Equation 1**.

### The Median Mystery

Now, let us use the second piece of information: the median is $45$. Since $45$ lies between $40$ and $50$, the class interval $40-50$ is our **Median Class**.

To apply the median formula, we identify the necessary parameters:
*   $L$ (lower limit of median class) $= 40$
*   $N/2 = 584 / 2 = 292$
*   $cf$ (cumulative frequency before median class) $= \alpha + 110 + 54 = \alpha + 164$
*   $f$ (frequency of median class) $= 30$
*   $h$ (class width) $= 10$

### The Algebra of Discovery

The standard formula for the median of grouped data is:
$$M = L + \left( \frac{\frac{N}{2} - cf}{f} ight) 	imes h$$

Substituting our known values into the formula:
$$45 = 40 + \left( \frac{292 - (\alpha + 164)}{30} ight) 	imes 10$$

Subtracting $40$ from both sides and simplifying the fraction:
$$5 = \left( \frac{128 - \alpha}{3} ight)$$

Multiplying by $3$ yields $15 = 128 - \alpha$. Therefore:
$$\alpha = 128 - 15 = 113$$

### Final Calculation

Now that we have $\alpha = 113$, we return to **Equation 1** to find $\beta$:
$$113 + \beta = 390$$
$$\beta = 390 - 113 = 277$$

The question asks for the absolute difference between $\alpha$ and $\beta$:
$$| \alpha - \beta | = | 113 - 277 |$$
$$| -164 | = 164$$

The final answer is **164**.

Animated Solution for Mathematics - Statistics: Consider the following frequency distribution $c l a ss : F r e q u e n cy : 10 - 20 α 20 - 30 110 30 - 40 54 40 - 50 30 50 - 60 β$ If the sum of all frequencies is 584 and median is 45, then $∣ α - β ∣$ is equal to

Enter Numerical Value:

Visualized Solution

Analyzing the Frequency Table

Sum of Frequencies Equation

Simplifying for $α + β$

Identifying the Median Class

Calculating Cumulative Frequencies

The Median Formula

Defining the Variables

Substitution in Formula

Solving for $α$

Solving for $β$

Calculating $∣ α - β ∣$

The Sigma Insight: Measures of Central Tendency (Mean, Median, Mode)

Analyzing the Setup

The First Constraint

The Median Mystery

The Algebra of Discovery

Final Calculation

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