The sum of deviations of a set of n values measured from 50 is – 10 and the sum of deviations of the values measured from 46 is 70. What is the mean of the values ?

Question

The sum of deviations of a set of n values measured from 50 is &ndash; 10 and the sum of deviations of the values measured from 46 is 70. What is the mean of the values ?

Prepp Team · Accepted Answer

Calculating the Mean from Sum of Deviations

This question asks us to find the mean of a set of 'n' values given the sum of their deviations from two different constants. The sum of deviations is a key concept in statistics, closely related to the mean.

Let the set of 'n' values be \(x_1, x_2, \ldots, x_n\). The mean of these values is denoted by \(\bar{x}\).

The sum of deviations of the values from a constant 'a' is given by the formula:

\(\sum_{i=1}^{n} (x_i - a)\)

We can expand this sum:

\(\sum_{i=1}^{n} (x_i - a) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} a\)

Since \(\sum_{i=1}^{n} x_i = n\bar{x}\) (the sum of values equals the number of values multiplied by their mean) and \(\sum_{i=1}^{n} a = n \cdot a\) (summing a constant 'n' times), the formula becomes:

\(\sum_{i=1}^{n} (x_i - a) = n\bar{x} - na = n(\bar{x} - a)\)

We are given two pieces of information about the sum of deviations:

The sum of deviations measured from 50 is -10.
The sum of deviations measured from 46 is 70.

Using the formula \( \sum (x_i - a) = n(\bar{x} - a) \), we can write these as two equations:

Sum of deviations from 50: \( n(\bar{x} - 50) = -10 \)
Sum of deviations from 46: \( n(\bar{x} - 46) = 70 \)

We now have a system of two equations with two unknowns, \(n\) and \(\bar{x}\). We can solve for \(\bar{x}\).

From equation (1), divide both sides by \(n\):

\(\bar{x} - 50 = -\frac{10}{n}\) \(\bar{x} = 50 - \frac{10}{n} \quad \ldots (3)\)

From equation (2), divide both sides by \(n\):

\(\bar{x} - 46 = \frac{70}{n}\) \(\bar{x} = 46 + \frac{70}{n} \quad \ldots (4)\)

Now, we can equate the expressions for \(\bar{x}\) from equations (3) and (4):

\(50 - \frac{10}{n} = 46 + \frac{70}{n}\)

Rearrange the terms to solve for \(n\):

\(50 - 46 = \frac{70}{n} + \frac{10}{n}\) \(4 = \frac{80}{n}\)

Multiply both sides by \(n\) and divide by 4:

\(4n = 80\) \(n = \frac{80}{4}\) \(n = 20\)

So, there are 20 values in the set. Now we can substitute the value of \(n\) into either equation (3) or (4) to find the mean \(\bar{x}\).

Using equation (3):

\(\bar{x} = 50 - \frac{10}{n}\) \(\bar{x} = 50 - \frac{10}{20}\) \(\bar{x} = 50 - 0.5\) \(\bar{x} = 49.5\)

Alternatively, using equation (4):

\(\bar{x} = 46 + \frac{70}{n}\) \(\bar{x} = 46 + \frac{70}{20}\) \(\bar{x} = 46 + 3.5\) \(\bar{x} = 49.5\)

Both methods give the same mean value.

The mean of the values is 49.5.

Concept	Formula
Deviation of \(x_i\) from a constant 'a'	\(x_i - a\)
Sum of Deviations from 'a'	\( \sum (x_i - a) \)
Relation between Sum of Deviations, Mean (\(\bar{x}\)) and constant (a)	\( \sum (x_i - a) = n(\bar{x} - a) \)

Revision Table: Key Statistics Formulas

Term	Definition/Formula
Mean (\(\bar{x}\))	The average of a set of values. \( \bar{x} = \frac{\sum x_i}{n} \)
Deviation	The difference between a value and a reference point (often the mean or a constant).
Sum of Deviations from Mean	The sum of differences between each value and the mean. This sum is always zero: \( \sum (x_i - \bar{x}) = 0 \)
Sum of Deviations from Constant (a)	\( \sum (x_i - a) = n(\bar{x} - a) \)

Additional Information on Mean and Deviations

The mean is a measure of central tendency, representing the typical value in a dataset. Deviations from the mean are crucial for understanding the spread or variability of data, leading to concepts like variance and standard deviation.

When calculating the sum of deviations from a constant 'a', the result \( n(\bar{x} - a) \) shows that the sum is directly proportional to the difference between the mean (\(\bar{x}\)) and the constant (a), and the number of observations (n). If 'a' is equal to the mean, the sum of deviations is \(n(\bar{x} - \bar{x}) = n \cdot 0 = 0\), which confirms the property that the sum of deviations from the mean is always zero.

In this problem, we used the property \( \sum (x_i - a) = n(\bar{x} - a) \) with two different values of 'a' to create a system of equations that allowed us to solve for the mean without explicitly knowing all the values \(x_i\) or the number of values \(n\) beforehand, although we found \(n\) during the process.