Calculating Sum of Deviations from Median for AP Marks

The problem provides data about the marks obtained by 51 students. These marks form an Arithmetic Progression (AP) with the first term \($a = 4$\) and a common difference \($d = 3$\). The total number of students, and thus the number of terms in the AP, is \($n = 51$\).

We are asked to find the sum of the deviations of these marks from the median.

Let the marks be denoted by \($x_1, x_2, \ldots, x_{51}$\). The median is the middle value of the dataset when arranged in order. Since the data is already in an AP, it is ordered. For an odd number of terms (\($n=51$\)), the median is the term at the \(\frac{n+1}{2}\)-th position.

• Number of terms, \($n = 51$\).
• Position of the median term \(= \frac{51+1}{2} = \frac{52}{2} = 26\)-th term.

Let the median be \(M\). The median is the 26th term of the AP. The formula for the \(k\)-th term of an AP is \($a_k = a + (k-1)d$\).

• The median term \(M = a_{26} = a + (26-1)d\).
• \(M = 4 + (25) \times 3\).
• \(M = 4 + 75\).
• \(M = 79\).

The median mark is 79.

We need to find the sum of the deviations from the median, which is given by \(\sum_{i=1}^{51} (x_i - M)\). This sum can be expanded as:

\(\sum_{i=1}^{51} (x_i - M) = \sum_{i=1}^{51} x_i - \sum_{i=1}^{51} M\)

\(\sum_{i=1}^{51} M = 51 \times M = 51 \times 79\)

So, the sum of deviations is \(\sum_{i=1}^{51} x_i - 51 \times 79\).

Now let's consider the properties of an Arithmetic Progression and statistical measures.

For a dataset that is symmetric, the mean and the median are equal. An Arithmetic Progression with an odd number of terms is a symmetric distribution.

• The mean of the AP is \(\bar{x} = \frac{\text{Sum of terms}}{\text{Number of terms}} = \frac{\sum x_i}{n}\).
• The sum of deviations from the mean, \(\sum_{i=1}^{n} (x_i - \bar{x})\), is always zero.

In this case, since the dataset is an AP with an odd number of terms, the mean is equal to the median. We calculated the median \(M = 79\). Let's calculate the mean to confirm.

The sum of an AP is \($S_n = \frac{n}{2}(a_1 + a_n)$\). We need the last term, \(a_{51}\).

• \(a_{51} = a + (51-1)d = 4 + 50 \times 3 = 4 + 150 = 154\).
• Sum of marks, \(S_{51} = \frac{51}{2}(4 + 154) = \frac{51}{2}(158) = 51 \times 79\).

The mean \(\bar{x} = \frac{S_{51}}{51} = \frac{51 \times 79}{51} = 79\).

As expected, the mean (\(\bar{x} = 79\)) is equal to the median (\(M = 79\)).

Since the mean and the median are equal (\(\bar{x} = M\)), the sum of deviations from the median is the same as the sum of deviations from the mean.

\(\sum_{i=1}^{51} (x_i - M) = \sum_{i=1}^{51} (x_i - \bar{x})\)

We know that the sum of deviations from the mean is always zero.

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

Since \(\bar{x} = \frac{\sum x_i}{n}\), then \(n\bar{x} = \sum x_i\).

Therefore, \(\sum_{i=1}^{n} (x_i - \bar{x}) = \sum x_i - \sum x_i = 0\).

Thus, the sum of the deviations measured from the median for this dataset is 0.

Revision Table: Key Concepts

Additional Information on Deviations and Central Tendency

Understanding deviations from central tendency measures is important in statistics. A deviation is simply the difference between an individual data point and a reference point (like the mean or median).

• Sum of Deviations from the Mean: A fundamental property of the arithmetic mean (\(\bar{x}\)) is that the sum of the deviations of all observations from the mean is always zero. This indicates that the mean is the balancing point of the dataset.
• Sum of Absolute Deviations from the Median: The sum of the absolute deviations (\(\sum |x_i - M|\)) is minimized when the reference point \(M\) is the median. This is a key property used in defining the median as a measure of central tendency that is resistant to extreme values (outliers).
• Sum of Squared Deviations from the Mean: The sum of the squared deviations (\(\sum (x_i - \bar{x})^2\)) is minimized when the reference point is the mean. This property is the basis for calculations like variance and standard deviation.

In the specific case of a perfectly symmetric distribution, the mean and median coincide. When the mean and median are the same, the sum of deviations from the median becomes equal to the sum of deviations from the mean, which is zero.

Since the marks form an AP with an odd number of terms, the distribution of marks is symmetric around the middle term (the median/mean). Therefore, the sum of the deviations from the median is zero.