The sum of deviations of n numbers from 10 and 20 are a, b respectively. If \(\frac{b}{a}\) = -4, then what is the mean of these n numbers ?
12
The problem asks us to find the mean of \(n\) numbers given information about the sum of their deviations from two different values, 10 and 20. We are given that the sum of deviations from 10 is \(a\), the sum of deviations from 20 is \(b\), and the ratio \(b/a\) is -4.
Understanding Sum of Deviations
The deviation of a number \(x_i\) from a constant value \(c\) is \(x_i - c\). The sum of deviations of \(n\) numbers \(x_1, x_2, \dots, x_n\) from a constant \(c\) is given by \(\sum_{i=1}^{n} (x_i - c)\).
Setting up the Equations for Sums of Deviations
Let the \(n\) numbers be \(x_1, x_2, \dots, x_n\). The sum of deviations from 10 is \(a\):
\[ a = \sum_{i=1}^{n} (x_i - 10) \]We can expand this sum:
\[ a = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} 10 \] \[ a = \sum_{i=1}^{n} x_i - 10n \]The sum of deviations from 20 is \(b\):
\[ b = \sum_{i=1}^{n} (x_i - 20) \]Expanding this sum:
\[ b = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} 20 \] \[ b = \sum_{i=1}^{n} x_i - 20n \]Relating Sums to the Mean
Let the mean of the \(n\) numbers be \(\bar{x}\). The mean is defined as:
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]From this definition, the sum of the numbers can be expressed in terms of the mean and \(n\):
\[ \sum_{i=1}^{n} x_i = n\bar{x} \]Now, substitute this expression for \(\sum_{i=1}^{n} x_i\) into the equations for \(a\) and \(b\):
For \(a\):
\[ a = n\bar{x} - 10n = n(\bar{x} - 10) \]For \(b\):
\[ b = n\bar{x} - 20n = n(\bar{x} - 20) \]Using the Given Ratio to Find the Mean
We are given the condition \(b/a = -4\). Substitute the expressions for \(a\) and \(b\) in terms of \(\bar{x}\) and \(n\) into this ratio:
\[ \frac{n(\bar{x} - 20)}{n(\bar{x} - 10)} = -4 \]Assuming \(n \ne 0\) (since we have \(n\) numbers), we can cancel \(n\) from the numerator and denominator:
\[ \frac{\bar{x} - 20}{\bar{x} - 10} = -4 \]Now, we solve this equation for \(\bar{x}\). Multiply both sides by \((\bar{x} - 10)\):
\[ \bar{x} - 20 = -4(\bar{x} - 10) \]Distribute -4 on the right side:
\[ \bar{x} - 20 = -4\bar{x} + 40 \]Collect the \(\bar{x}\) terms on one side and the constant terms on the other. Add \(4\bar{x}\) to both sides:
\[ \bar{x} + 4\bar{x} - 20 = 40 \] \[ 5\bar{x} - 20 = 40 \]Add 20 to both sides:
\[ 5\bar{x} = 40 + 20 \] \[ 5\bar{x} = 60 \]Divide by 5 to find \(\bar{x}\):
\[ \bar{x} = \frac{60}{5} \] \[ \bar{x} = 12 \]So, the mean of these \(n\) numbers is 12.
Verification
If the mean is 12, then:
- \(a = n(12 - 10) = n(2) = 2n\)
- \(b = n(12 - 20) = n(-8) = -8n\)
The ratio \(b/a\) would be \((-8n) / (2n)\). If \(n \ne 0\), this simplifies to \(-8/2 = -4\), which matches the given condition. This confirms our calculated mean is correct.
The mean of the \(n\) numbers is 12.
Revision Table: Key Concepts
| Concept | Definition/Formula | Application in Problem |
|---|---|---|
| Mean (\(\bar{x}\)) | Sum of numbers divided by count: \(\frac{\sum x_i}{n}\) | We used \(\sum x_i = n\bar{x}\) to express sums of deviations. |
| Deviation | Difference between a number and a constant: \(x_i - c\) | Used to define \(a\) and \(b\). |
| Sum of Deviations | \(\sum_{i=1}^{n} (x_i - c) = \sum x_i - nc\) | Formulas for \(a\) and \(b\). |
| Algebraic Manipulation | Solving equations for an unknown variable. | Used to solve the equation \(\frac{\bar{x} - 20}{\bar{x} - 10} = -4\) for \(\bar{x}\). |
Additional Information: Properties of Mean
The mean is a measure of central tendency. It has several useful properties:
- The sum of deviations of a set of numbers from their mean is always zero. That is, \(\sum_{i=1}^{n} (x_i - \bar{x}) = 0\).
Let's see why: \(\sum(x_i - \bar{x}) = \sum x_i - \sum \bar{x} = n\bar{x} - n\bar{x} = 0\). - The formula for the sum of deviations from any arbitrary constant \(c\) is \(\sum (x_i - c) = \sum x_i - nc\). Using \(\sum x_i = n\bar{x}\), this becomes \(\sum (x_i - c) = n\bar{x} - nc = n(\bar{x} - c)\).
This is exactly the formula we used for \(a\) and \(b\).
\(a = \sum (x_i - 10) = n(\bar{x} - 10)\)
\(b = \sum (x_i - 20) = n(\bar{x} - 20)\) - The mean is sensitive to extreme values (outliers).
- The mean is unique for any given set of data.
This problem is a good example of how the property \(\sum (x_i - c) = n(\bar{x} - c)\) can be used to relate sums of deviations from different points to the mean of the data.
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