Understanding the Problem: Frequency Distribution and Mean

The problem provides a frequency distribution table with class intervals and their corresponding frequencies. Some frequencies are expressed in terms of variables, 'p' and 'q'. We are given two crucial pieces of information: the total frequency and the mean of the distribution. Our goal is to use this information to find the values of 'p' and 'q' and then specifically determine the value of 'q'.

A frequency distribution organizes data into classes and shows how many observations fall into each class. The mean of a grouped frequency distribution is calculated using the formula:

\[ \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \]

Where:

• \(f_i\) is the frequency of the i-th class.
• \(x_i\) is the class mark (midpoint) of the i-th class.
• \(\sum f_i\) is the total frequency.
• \(\sum f_i x_i\) is the sum of the product of frequencies and class marks.

Step-by-Step Solution to Find the Value of q

1. Set up an Equation using the Total Frequency

The total frequency is given as 120. The sum of all frequencies in the table must equal this total.

\[ \text{Sum of Frequencies} = 17 + (p+q) + 32 + (p-3q) + 19 = 120 \]

Combining like terms:

\[ (p+p) + (q-3q) + (17+32+19) = 120 \]

\[ 2p - 2q + 68 = 120 \]

\[ 2p - 2q = 120 - 68 \]

\[ 2p - 2q = 52 \]

Dividing the equation by 2:

\[ p - q = 26 \quad \text{(Equation 1)} \]

2. Calculate Class Marks (Midpoints)

To calculate the mean, we need the midpoint of each class interval. The class mark \(x_i\) is the average of the lower and upper limits of the class interval.

• Class 0-20: \(x_1 = (0 + 20) / 2 = 10\)
• Class 20-40: \(x_2 = (20 + 40) / 2 = 30\)
• Class 40-60: \(x_3 = (40 + 60) / 2 = 50\)
• Class 60-80: \(x_4 = (60 + 80) / 2 = 70\)
• Class 80-100: \(x_5 = (80 + 100) / 2 = 90\)

3. Calculate the Product of Frequency and Class Mark (\(f_i x_i\)) for each Class

Now, we multiply the frequency of each class by its corresponding class mark.

• Class 1: \(f_1 x_1 = 17 \times 10 = 170\)
• Class 2: \(f_2 x_2 = (p+q) \times 30 = 30p + 30q\)
• Class 3: \(f_3 x_3 = 32 \times 50 = 1600\)
• Class 4: \(f_4 x_4 = (p-3q) \times 70 = 70p - 210q\)
• Class 5: \(f_5 x_5 = 19 \times 90 = 1710\)

4. Calculate the Sum of \(f_i x_i\)

Add up the \(f_i x_i\) values for all classes:

\[ \sum f_i x_i = 170 + (30p + 30q) + 1600 + (70p - 210q) + 1710 \]

Combine terms with 'p', 'q', and constant terms:

\[ \sum f_i x_i = (30p + 70p) + (30q - 210q) + (170 + 1600 + 1710) \]

\[ \sum f_i x_i = 100p - 180q + 3480 \]

5. Set up an Equation using the Mean Formula

The mean is given as 50, and the total frequency (\(\sum f_i\)) is 120. Substitute these values and the expression for \(\sum f_i x_i\) into the mean formula:

\[ 50 = \frac{100p - 180q + 3480}{120} \]

Multiply both sides by 120:

\[ 50 \times 120 = 100p - 180q + 3480 \]

\[ 6000 = 100p - 180q + 3480 \]

Subtract 3480 from both sides:

\[ 6000 - 3480 = 100p - 180q \]

\[ 2520 = 100p - 180q \]

Divide the equation by the greatest common divisor, which is 20:

\[ \frac{2520}{20} = \frac{100p}{20} - \frac{180q}{20} \]

\[ 126 = 5p - 9q \quad \text{(Equation 2)} \]

6. Solve the System of Linear Equations

We now have a system of two linear equations with two variables, p and q:

• Equation 1: \(p - q = 26\)
• Equation 2: \(5p - 9q = 126\)

From Equation 1, we can express p in terms of q:

\[ p = q + 26 \]

Substitute this expression for p into Equation 2:

\[ 5(q + 26) - 9q = 126 \]

\[ 5q + 130 - 9q = 126 \]

Combine like terms:

\[ (5q - 9q) + 130 = 126 \]

\[ -4q + 130 = 126 \]

Subtract 130 from both sides:

\[ -4q = 126 - 130 \]

\[ -4q = -4 \]

Divide by -4:

\[ q = \frac{-4}{-4} \]

\[ q = 1 \]

7. Determine the Value of p (Optional Verification)

Although not asked, we can find the value of p using \(p = q + 26\):

\[ p = 1 + 26 = 27 \]

Now, let's check the frequencies with p=27 and q=1:

• 17
• \(p+q = 27+1 = 28\)
• 32
• \(p-3q = 27-3(1) = 27-3 = 24\)
• 19

The frequencies are 17, 28, 32, 24, 19. All are non-negative, which is good. Their sum is \(17+28+32+24+19 = 120\), which matches the given total frequency.

8. Summary Table

Here is the frequency distribution with class marks and \(f_i x_i\) expressions/values:

Using the calculated values p=27 and q=1, we can also fill in the actual frequencies and \(f_i x_i\) values:

Mean = \(\frac{6000}{120} = 50\), which matches the given information.

Final Answer Determination

From our calculations solving the system of equations, we found that the value of q is 1.

Revision Table: Key Concepts

Additional Information: Solving Systems of Equations

In this problem, finding the values of 'p' and 'q' required solving a system of two linear equations. A system of linear equations is a set of two or more linear equations involving the same variables.

Common methods for solving systems of two linear equations include:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This is the method used in the solution above.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable and solve for the remaining one.
• Graphical Method: Graph both equations on the same coordinate plane. The point where the lines intersect is the solution to the system. This method can be less precise for non-integer solutions.

Understanding how to set up and solve systems of equations is fundamental in various quantitative problems, including those in statistics where unknown frequencies or values need to be determined based on given statistical measures.