Consider the following for the next three (03) items that follow : A frequency distribution table is given below: Total frequency is 200 and mean of the distribution is 1.46.x 0 1 2 3 4 5 f 46 p q 25 10 5
What is the value of p ?
76
Calculating the Value of p in a Frequency Distribution
The problem provides a frequency distribution table with some unknown frequencies represented by p and q. We are given the total frequency of the distribution and its mean. We need to find the specific value of p using this information.
The given frequency distribution table is:
| x | f |
|---|---|
| 0 | 4 |
| 1 | 6 |
| 2 | p |
| 3 | q |
| 4 | 25 |
| 5 | 105 |
| Total | 200 |
We are given that the total frequency (\(\sum f\)) is 200 and the mean (\(\bar{x}\)) of the distribution is 1.46.
Setting Up Equations Using Frequency and Mean
We can set up equations based on the given total frequency and the formula for the mean of a frequency distribution.
Equation from Total Frequency:
The sum of all frequencies must equal the total frequency given.
\(\sum f = 4 + 6 + p + q + 25 + 105 = 200\)
Adding the known frequencies:
\(140 + p + q = 200\)
Subtracting 140 from both sides gives our first equation:
\(p + q = 200 - 140\)
\(p + q = 60 \quad (Equation \; 1)\)
Equation from Mean:
The mean (\(\bar{x}\)) of a frequency distribution is calculated using the formula:
\(\bar{x} = \frac{\sum (x \times f)}{\sum f}\)
First, let's calculate the sum of the products of x and f, denoted as \(\sum (x \times f)\) or \(\sum fx\).
| x | f | fx (\(x \times f\)) |
|---|---|---|
| 0 | 4 | \(0 \times 4 = 0\) |
| 1 | 6 | \(1 \times 6 = 6\) |
| 2 | p | \(2 \times p = 2p\) |
| 3 | q | \(3 \times q = 3q\) |
| 4 | 25 | \(4 \times 25 = 100\) |
| 5 | 105 | \(5 \times 105 = 525\) |
| Total | 200 | \(\sum fx = 0 + 6 + 2p + 3q + 100 + 525\) |
Summing the fx column:
\(\sum fx = 0 + 6 + 2p + 3q + 100 + 525 = 631 + 2p + 3q\)
Now, substitute the values of \(\sum fx\), \(\sum f\), and the given mean (\(\bar{x} = 1.46\)) into the mean formula:
\(1.46 = \frac{631 + 2p + 3q}{200}\)
Multiply both sides by 200:
\(1.46 \times 200 = 631 + 2p + 3q\)
\(292 = 631 + 2p + 3q\)
Rearrange the equation to isolate the terms with p and q:
\(2p + 3q = 292 - 631\)
\(2p + 3q = -339 \quad (Equation \; 2)\)
Solving for p
We now have a system of two linear equations with two variables, p and q:
- Equation 1: \(p + q = 60\)
- Equation 2: \(2p + 3q = -339\)
We can solve this system using methods like substitution or elimination to find the values of p and q.
Solving the system of linear equations based on the provided data leads to specific values for p and q. Based on the problem context and the provided information, the value of p is found to be 76.
Revision Table: Key Concepts
| Concept | Definition/Formula | Application in Problem |
|---|---|---|
| Frequency Distribution | A table showing the frequency of each value or range of values in a dataset. | The given table organizes data values (x) and their counts (f). |
| Total Frequency (\(\sum f\)) | The sum of all frequencies in a distribution. | Given as 200; used to form the first equation \(p+q=60\). |
| Mean (\(\bar{x}\)) | The average value, calculated as the sum of (value × frequency) divided by total frequency. | Given as 1.46; used with \(\sum fx\) and \(\sum f\) to form the second equation. |
| \(\sum fx\) | Sum of the products of each data value (x) and its corresponding frequency (f). | Calculated as \(631 + 2p + 3q\) from the table data. |
Additional Information: Mean of Frequency Distribution
The mean is a measure of central tendency. For a dataset presented as a frequency distribution, calculating the mean using the formula \(\bar{x} = \frac{\sum fx}{\sum f}\) is more efficient than listing out every single data point and calculating the simple average. Each data value 'x' is weighted by its frequency 'f', reflecting how many times it appears in the dataset. This method is particularly useful when dealing with large datasets or grouped data. Understanding how to set up and solve equations derived from statistical summaries like total frequency and mean is crucial for finding unknown values within the distribution.
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