Problem Breakdown:

The provided graph represents a frequency distribution, where the frequencies for two intervals (20–40 and 100–120) are missing. We are tasked with finding the values of these missing frequencies ($f_1$ and $f_2$), given the following:

1. Mean of the data: 57.6
2. Sum of all frequencies: 50.

Solution Approach:

Step 1: Identify Class Intervals and Given Frequencies

From the graph, we observe the class intervals and their frequencies:

• $0\text{–}20$: $6$
• $20\text{–}40$: $f_1$ (missing frequency)
• $40\text{–}60$: $12$
• $60\text{–}80$: $16$
• $80\text{–}100$: $7$
• $100\text{–}120$: $f_2$ (missing frequency)
• $120\text{–}140$: $4$

Step 2: Frequency Total Condition

The total sum of frequencies is 50: $6 + f_1 + 12 + 16 + 7 + f_2 + 4 = 50$ Simplifying: $45 + f_1 + f_2 = 50$ f_1 + f_2 = 5 \tag{1}

Step 3: Use the Mean Formula

The formula for the mean is: $\text{Mean} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}$ Here:

• $\sum f_i = 50$ (total frequency),
• $\text{Mean} = 57.6$,
• $x_i$ represents the midpoint of each class interval.

We calculate the midpoints ($x_i$) of each class:

• $0\text{–}20$: $x_1 = 10$,
• $20\text{–}40$: $x_2 = 30$,
• $40\text{–}60$: $x_3 = 50$,
• $60\text{–}80$: $x_4 = 70$,
• $80\text{–}100$: $x_5 = 90$,
• $100\text{–}120$: $x_6 = 110$,
• $120\text{–}140$: $x_7 = 130$.

The mean equation becomes: $57.6 = \frac{(10 \cdot 6) + (30 \cdot f_1) + (50 \cdot 12) + (70 \cdot 16) + (90 \cdot 7) + (110 \cdot f_2) + (130 \cdot 4)}{50}$ Simplify the known terms: $57.6 = \frac{60 + 30f_1 + 600 + 1120 + 630 + 110f_2 + 520}{50}$ $57.6 = \frac{2930 + 30f_1 + 110f_2}{50}$ Multiply through by 50: $2880 = 2930 + 30f_1 + 110f_2$ Simplify: -50 = 30f_1 + 110f_2 \tag{2}

Step 4: Solve Simultaneous Equations

From (1): f_1 + f_2 = 5 \tag{1} From (2): 30f_1 + 110f_2 = -50 \tag{2}

Substitute $f_2 = 5 - f_1$ from (1) into (2): $30f_1 + 110(5 - f_1) = -50$ $30f_1 + 550 - 110f_1 = -50$ $-80f_1 + 550 = -50$ $-80f_1 = -600$ $f_1 = 7.5$

Substitute $f_1 = 7.5$ into $f_1 + f_2 = 5$: $7.5 + f_2 = 5$ $f_2 = -2.5$

Correct if Need Adjustment (Double Check Frequencies match)>Let's revisit the frequency distribution problem systematically to correct the issue from Step 4:

Step 4: Re-solve Simultaneous Equations

We had these two equations from earlier:

1. f_1 + f_2 = 5 \tag{1}
2. 30f_1 + 110f_2 = -50 \tag{2}

Substitute $f_2 = 5 - f_1$ (from Equation 1) into Equation 2:

$30f_1 + 110(5 - f_1) = -50$

Expand:

$30f_1 + 550 - 110f_1 = -50$

Simplify:

$-80f_1 + 550 = -50$

$-80f_1 = -600$

$f_1 = 7.5$

Find $f_2$:

Substitute $f_1 = 7.5$ into $f_1 + f_2 = 5$:

$7.5 + f_2 = 5$

$f_2 = -2.5$

Error Detected: Negative Frequency

This result does not make sense because frequencies cannot be negative. There seems to be an issue in the constraints or initial assumptions. Let’s carefully re-check the calculations or clarify the problem’s constraints. Would you like to go through this systematically to identify the issue?