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.

1. 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).

2. 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).

3. 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: \(n\bar{x} + 3n = 54 \implies n(\bar{x} + 3) = 54\).

4. 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.