If X̅ = 20 is the mean of 10 observations x1, x2, ... x10; then what is the value of \(\displaystyle \sum_{i=1}^{10}\left(\frac{3 x_i-4}{5}\right) ?\) ?
112
The problem provides the mean of 10 observations and asks for the value of a specific summation involving a linear transformation of these observations. We need to use the properties of summation and the definition of the mean to solve this statistics problem.
We are given:
The mean (\(\bar{X}\)) of a set of observations is defined as the sum of the observations divided by the number of observations. Mathematically, for 10 observations, this is:
\(\bar{X} = \frac{\sum_{i=1}^{10} x_i}{10}\)
We are given \(\bar{X} = 20\), so we can write:
\(20 = \frac{\sum_{i=1}^{10} x_i}{10}\)
To find the sum of the observations (\(\sum_{i=1}^{10} x_i\)), we can multiply both sides by 10:
\(\sum_{i=1}^{10} x_i = 20 \times 10 = 200\)
So, the sum of the original 10 observations is 200.
We need to find the value of the following summation:
\(\sum_{i=1}^{10}\left(\frac{3 x_i-4}{5}\right)\)
This expression involves a linear transformation of each observation \(x_i\), specifically \(y_i = \frac{3 x_i - 4}{5}\). We can use the properties of summation to simplify this expression. The summation of a linear transformation \(ax_i + b\) can be written as \(a \sum x_i + \sum b\).
Let's rewrite the expression inside the summation:
\(\frac{3 x_i-4}{5} = \frac{3}{5} x_i - \frac{4}{5}\)
Now, apply the summation property:
\(\sum_{i=1}^{10}\left(\frac{3}{5} x_i - \frac{4}{5}\right) = \sum_{i=1}^{10}\left(\frac{3}{5} x_i\right) + \sum_{i=1}^{10}\left(-\frac{4}{5}\right)\)
The property \(\sum (ax_i) = a \sum x_i\) allows us to pull the constant \(\frac{3}{5}\) out of the first summation:
\(= \frac{3}{5} \sum_{i=1}^{10} x_i + \sum_{i=1}^{10}\left(-\frac{4}{5}\right)\)
The sum of a constant over \(n\) terms is simply \(n\) times the constant. Here, the constant is \(-\frac{4}{5}\) and the number of terms is 10:
\(\sum_{i=1}^{10}\left(-\frac{4}{5}\right) = 10 \times \left(-\frac{4}{5}\right) = -\frac{40}{5} = -8\)
Now substitute this back into the expression, along with the sum of \(x_i\) we calculated earlier (\(\sum_{i=1}^{10} x_i = 200\)):
\(= \frac{3}{5} \times (200) + (-8)\)
Perform the multiplication:
\(= 3 \times \left(\frac{200}{5}\right) - 8\)
\(= 3 \times 40 - 8\)
\(= 120 - 8\)
\(= 112\)
The value of the given summation \(\displaystyle \sum_{i=1}^{10}\left(\frac{3 x_i-4}{5}\right)\) is 112.
| Step | Description | Calculation |
|---|---|---|
| 1 | Find the sum of observations (\(\sum x_i\)) using the given mean. | \(\sum_{i=1}^{10} x_i = \bar{X} \times n = 20 \times 10 = 200\) |
| 2 | Rewrite the expression inside the summation. | \(\frac{3 x_i-4}{5} = \frac{3}{5} x_i - \frac{4}{5}\) |
| 3 | Apply summation properties. | \(\sum_{i=1}^{10}\left(\frac{3}{5} x_i - \frac{4}{5}\right) = \frac{3}{5}\sum x_i + \sum \left(-\frac{4}{5}\right)\) |
| 4 | Calculate the sum of the constant term. | \(\sum_{i=1}^{10}\left(-\frac{4}{5}\right) = 10 \times \left(-\frac{4}{5}\right) = -8\) |
| 5 | Substitute values and compute. | \(\frac{3}{5} \times 200 - 8 = 120 - 8 = 112\) |
| Concept | Definition/Formula |
|---|---|
| Mean (\(\bar{X}\)) | Sum of observations divided by the number of observations: \(\bar{X} = \frac{\sum x_i}{n}\) |
| Summation (\(\sum\)) | Operator representing the sum of a sequence of terms. |
| Summation Property (Linearity) | \(\sum_{i=1}^{n} (a x_i + b) = a \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} b\) |
| Sum of a Constant | \(\sum_{i=1}^{n} c = n \times c\) |
When a dataset \(x_1, x_2, \ldots, x_n\) is transformed linearly to \(y_i = ax_i + b\), the sum of the transformed data is related to the sum of the original data and the constants \(a\) and \(b\). This is exactly what we used in this problem:
\(\sum_{i=1}^{n} y_i = \sum_{i=1}^{n} (ax_i + b) = a \sum_{i=1}^{n} x_i + nb\)
This property is very useful in statistics, especially when dealing with changes in scale and origin of data. For example, if you convert temperatures from Celsius to Fahrenheit, you are performing a linear transformation, and this property helps relate the sum (or mean) of temperatures in one scale to the sum (or mean) in the other scale.
The mean of the transformed data (\(\bar{Y}\)) is also linearly related to the mean of the original data (\(\bar{X}\)):
\(\bar{Y} = \frac{\sum y_i}{n} = \frac{a \sum x_i + nb}{n} = \frac{a \sum x_i}{n} + \frac{nb}{n} = a \bar{X} + b\)
In our specific problem, the transformation is \(y_i = \frac{3}{5} x_i - \frac{4}{5}\), with \(a = \frac{3}{5}\) and \(b = -\frac{4}{5}\). The mean of the transformed data would be:
\(\bar{Y} = \frac{3}{5} \bar{X} - \frac{4}{5} = \frac{3}{5}(20) - \frac{4}{5} = 12 - 0.8 = 11.2\)
The sum we calculated (\(112\)) is actually \(n \times \bar{Y} = 10 \times 11.2 = 112\), which confirms our result using the property for the mean of transformed data as well.
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