Given two sets:

• Set \( A \) has median \( m_1 = 2 \)
• Set \( B \) has median \( m_2 = 4 \)

What can we say about the median of the combined set \( A \cup B \)?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

\[ \text{Combined Median} \in [2, 4] \]

So, the exact median cannot be determined with the given data.

Mean, Median, and Mode Relation

Given:

• Mean = 1
• Median = \( 3^x \)
• Mode = \( 9^x \)

Use empirical formula:

\[ \text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean} \]

\[ 9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2 \]

Let \( y = 3^x \), then:

\[ y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0 \]

So, \( y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4 \)

✅ Final Answer: \(\boxed{1 \text{ or } 4}\)

Corrected Mean and Standard Deviation

Original Mean: 40, Standard Deviation: 15

Two scores were misread: 25 → 52 and 35 → 53

Corrected Mean:

\( \mu' = \frac{3955}{100} = \boxed{39.55} \)

Corrected Standard Deviation:

\( \sigma' = \sqrt{\frac{178837}{100} - (39.55)^2} \approx \boxed{14.96} \)

✅ Final Answer: Mean = 39.55, Standard Deviation ≈ 14.96

Median & Frequency Table

Given: Median = 42.6, Total Frequency = 685

Using Median Formula:

\( \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \cdot h \)

• Median Class: 40–50
• Lower boundary \( L = 40 \)
• Frequency \( f = 180 \)
• Class width \( h = 10 \)
• Cumulative freq before median class \( F = 214 + f_1 \)

Substituting values:

\( 42.6 = 40 + \left( \frac{128.5 - f_1}{180} \right) \cdot 10 \Rightarrow f_1 = \boxed{82} \)

Using total frequency:

\( 662 + f_2 = 685 \Rightarrow f_2 = \boxed{23} \)

✅ Final Answer: \( f_1 = 82,\quad f_2 = 23 \)