Consider the following frequency distribution for the next three (03) items follow : The total frequency is 120. The mean is 50.Class 0-20 20-40 40-60 60-80 80-100 Frequency 17 p + q 32 p - 3q 19
What is the value of p ?
27
Understanding Frequency Distribution and Mean Calculation
This problem involves a frequency distribution where some frequencies are unknown, represented by variables \(p\) and \(q\). We are given the total frequency and the mean of the distribution. Our goal is to use this information to find the value of \(p\).
Calculating Mid-points for Frequency Distribution
To calculate the mean of a grouped frequency distribution, we first need to find the mid-point (or class mark) for each class interval. The mid-point is the average of the lower and upper limits of a class.
| Class | Lower Limit | Upper Limit | Mid-point (\(x_i\)) |
|---|---|---|---|
| 0-20 | 0 | 20 | \(\frac{0+20}{2} = 10\) |
| 20-40 | 20 | 40 | \(\frac{20+40}{2} = 30\) |
| 40-60 | 40 | 60 | \(\frac{40+60}{2} = 50\) |
| 60-80 | 60 | 80 | \(\frac{60+80}{2} = 70\) |
| 80-100 | 80 | 100 | \(\frac{80+100}{2} = 90\) |
Setting Up Equations from Given Information
We are given two key pieces of information: the total frequency and the mean. We can use these to form equations involving \(p\) and \(q\).
Equation from Total Frequency
The sum of all frequencies must equal the total frequency given, which is 120.
Sum of frequencies \( = 17 + (p + q) + 32 + (p - 3q) + 19 = 120\)
Combining like terms:
\[(17 + 32 + 19) + (p + p) + (q - 3q) = 120\] \[68 + 2p - 2q = 120\] \[2p - 2q = 120 - 68\] \[2p - 2q = 52\]Dividing the entire equation by 2:
\[p - q = 26 \quad \text{(Equation 1)}\]Equation from Mean
The formula for the mean (\(\bar{x}\)) of a grouped frequency distribution is:
\[\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\]Where \(f_i\) is the frequency of each class and \(x_i\) is the mid-point of each class.
First, let's calculate the sum of the products of frequency and mid-point (\(\sum f_i x_i\)):
\[\sum f_i x_i = (17 \times 10) + ((p + q) \times 30) + (32 \times 50) + ((p - 3q) \times 70) + (19 \times 90)\] \[\sum f_i x_i = 170 + 30(p + q) + 1600 + 70(p - 3q) + 1710\] \[\sum f_i x_i = 170 + 30p + 30q + 1600 + 70p - 210q + 1710\]Combining like terms:
\[\sum f_i x_i = (30p + 70p) + (30q - 210q) + (170 + 1600 + 1710)\] \[\sum f_i x_i = 100p - 180q + 3480\]Now, substitute this into the mean formula. We are given that the mean is 50 and the total frequency (\(\sum f_i\)) is 120.
\[50 = \frac{100p - 180q + 3480}{120}\]Multiply both sides by 120:
\[50 \times 120 = 100p - 180q + 3480\] \[6000 = 100p - 180q + 3480\]Subtract 3480 from both sides:
\[6000 - 3480 = 100p - 180q\] \[2520 = 100p - 180q\]Dividing the entire equation by 20 to simplify:
\[\frac{2520}{20} = \frac{100p}{20} - \frac{180q}{20}\] \[126 = 5p - 9q \quad \text{(Equation 2)}\]Solving the System of Linear Equations
We now have a system of two linear equations with two variables:
- \(p - q = 26\)
- \(5p - 9q = 126\)
From Equation 1, we can express \(p\) in terms of \(q\):
\[p = q + 26\]Substitute this expression for \(p\) into Equation 2:
\[5(q + 26) - 9q = 126\] \[5q + 5 \times 26 - 9q = 126\] \[5q + 130 - 9q = 126\]Combine the terms with \(q\):
\[(5q - 9q) + 130 = 126\] \[-4q + 130 = 126\]Subtract 130 from both sides:
\[-4q = 126 - 130\] \[-4q = -4\]Divide by -4 to find the value of \(q\):
\[q = \frac{-4}{-4}\] \[q = 1\]Now that we have the value of \(q\), we can substitute it back into the expression for \(p\):
\[p = q + 26\] \[p = 1 + 26\] \[p = 27\]Thus, the value of \(p\) is 27.
Verification
Let's check if these values of \(p\) and \(q\) are consistent with the given total frequency and mean.
The frequencies are:
- 17
- \(p + q = 27 + 1 = 28\)
- 32
- \(p - 3q = 27 - 3(1) = 27 - 3 = 24\)
- 19
Sum of frequencies \( = 17 + 28 + 32 + 24 + 19 = 120\). This matches the given total frequency.
Now, let's calculate the mean with these frequencies:
\[\sum f_i x_i = (17 \times 10) + (28 \times 30) + (32 \times 50) + (24 \times 70) + (19 \times 90)\] \[\sum f_i x_i = 170 + 840 + 1600 + 1680 + 1710\] \[\sum f_i x_i = 6000\]Mean \( = \frac{6000}{120} = 50\). This matches the given mean.
The values \(p = 27\) and \(q = 1\) are correct.
Revision Table: Frequency Distribution Analysis
| Concept | Description | Application in Problem |
|---|---|---|
| Frequency Distribution | A table showing the frequency of occurrence of data within given intervals (classes). | Provided data with class intervals and frequencies (including unknown \(p, q\)). |
| Mid-point (Class Mark) | The average of the lower and upper limits of a class interval, used to represent the class in calculations. | Calculated for each class to find \(\sum f_i x_i\). |
| Total Frequency | The sum of all frequencies in a distribution. | Used to form the first equation: \( \sum f_i = 120 \). |
| Mean of Grouped Data | A measure of central tendency calculated as \( \frac{\sum f_i x_i}{\sum f_i} \). | Used to form the second equation: \( \frac{\sum f_i x_i}{120} = 50 \). |
| System of Linear Equations | A set of two or more linear equations with the same variables, solved simultaneously to find variable values. | Two equations with \(p\) and \(q\) were solved to find their values. |
Additional Information: Solving for Variables in Statistics Problems
Problems involving unknown frequencies or class limits often require setting up equations based on given statistical measures like mean, median, or mode, or total frequency. For frequency distributions:
- If the total frequency is given, sum up all frequency expressions and equate them to the total.
- If the mean is given, calculate \(\sum f_i x_i\) using mid-points and frequency expressions, and use the mean formula \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\).
- If the median or mode is given, you would use their respective formulas, which involve finding the median class or modal class first. These can be more complex if the unknown variables affect the location of these classes.
- For problems with two unknowns (like \(p\) and \(q\) here), you typically need two independent pieces of information (like total frequency and mean) to create a system of two equations that can be solved simultaneously.
- Solving systems of equations can be done using substitution (as shown in this solution) or elimination methods. Choose the method that seems easiest based on the equations formed.
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