3.75
Problem Analysis:
We are given the sum of deviations of a set of \(n\) numbers (\(x_1, x_2, ..., x_n\)) from two different points (15 and -3) and asked to find the arithmetic mean (\(\bar{x}\)).
- Let the set of numbers be \(x_1, x_2, ..., x_n\).
- The sum of deviations from 15 is \(\sum_{i=1}^{n} (x_i - 15) = -90\).
- The sum of deviations from -3 is \(\sum_{i=1}^{n} (x_i - (-3)) = \sum_{i=1}^{n} (x_i + 3) = 54\).
- We need to find the arithmetic mean, \(\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}\).
Arithmetic Mean Calculation Using Deviations
We can use the properties of summation to simplify the given information.
- Simplify the first condition:
\(\sum_{i=1}^{n} (x_i - 15) = -90\)
This expands to: \(\sum_{i=1}^{n} x_i - \sum_{i=1}^{n} 15 = -90\)
Which simplifies to: \(\sum x_i - 15n = -90\). Let's call this Equation (1).
- Simplify the second condition:
\(\sum_{i=1}^{n} (x_i + 3) = 54\)
This expands to: \(\sum_{i=1}^{n} x_i + \sum_{i=1}^{n} 3 = 54\)
Which simplifies to: \(\sum x_i + 3n = 54\). Let's call this Equation (2).
- Relate the sum of deviations to the arithmetic mean:
Recall that \(\sum x_i = n\bar{x}\). Substitute this into Equations (1) and (2).
Equation (1) becomes: \(n\bar{x} - 15n = -90 \implies n(\bar{x} - 15) = -90\).
Equation (2) becomes: \(n\bar{x} + 3n = 54 \implies n(\bar{x} + 3) = 54\).
- Solve for the arithmetic mean (\(\bar{x}\)):
We have a system of two equations. Divide the first equation by the second:
\(\frac{n(\bar{x} - 15)}{n(\bar{x} + 3)} = \frac{-90}{54}\)
Simplify the equation:
\(\frac{\bar{x} - 15}{\bar{x} + 3} = -\frac{5}{3}\)
Now, cross-multiply:
\(3(\bar{x} - 15) = -5(\bar{x} + 3)\)
\(3\bar{x} - 45 = -5\bar{x} - 15\)
Combine like terms:
\(3\bar{x} + 5\bar{x} = 45 - 15\)
\(8\bar{x} = 30\)
Solve for \(\bar{x}\):
\(\bar{x} = \frac{30}{8} = \frac{15}{4}\)
\(\bar{x} = 3.75\)
The arithmetic mean is 3.75.
Similar Questions
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