Question:medium

Find the mean deviation about the mean for the data set: 1, 3, 5, 7, \dots, 101

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For any arithmetic progression with $n$ terms and common difference $d$, if $n$ is odd, the mean deviation about the mean is $\frac{(n^2 - 1)d}{4n}$. Plugging in our values: $\frac{(51^2 - 1) \times 2}{4 \times 51} = \frac{2600 \times 2}{204} = \frac{5200}{204} \approx 25.49$.
Updated On: Apr 16, 2026
  • 24
  • 25
  • 25.5
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The mean deviation about the mean is a measure of dispersion. It is calculated by taking the average of the absolute differences between each data point and the mean of the data set.
Step 2: Key Formula or Approach:
The formula for Mean Deviation (MD) about the mean $\bar{x}$ is: \[ \text{MD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| \] First, we must find the total number of terms $n$ and the mean $\bar{x}$ for the given arithmetic progression (AP).
Step 3: Detailed Explanation:
The given sequence is an AP: $1, 3, 5, \dots, 101$.
First term $a = 1$, common difference $d = 2$.
To find the number of terms $n$: \[ n\text{-th term} = a + (n-1)d \implies 101 = 1 + (n-1)2 \] \[ 100 = (n-1)2 \implies n-1 = 50 \implies n = 51 \] Since it is an AP with an odd number of terms, the mean $\bar{x}$ is simply the middle term, or the average of the first and last terms: \[ \bar{x} = \frac{1 + 101}{2} = 51 \] Now, we calculate the absolute deviations $|x_i - \bar{x}|$ for each term: For $x_i = 1, 3, 5, \dots, 49, 51, 53, \dots, 99, 101$, the mean is $51$. Deviations are: $|1 - 51| = 50$
$|3 - 51| = 48$
$\dots$
$|49 - 51| = 2$
$|51 - 51| = 0$
$|53 - 51| = 2$
$\dots$
$|101 - 51| = 50$
The deviations form two identical sequences from $2$ to $50$ (even numbers). There are 25 such even numbers on either side of the mean. Sum of absolute deviations = $2 \times (2 + 4 + 6 + \dots + 50)$ \[ = 2 \times 2(1 + 2 + 3 + \dots + 25) = 4 \times \frac{25 \times (25+1)}{2} \] \[ = 4 \times \frac{25 \times 26}{2} = 2 \times 650 = 1300 \] Now, divide by the total number of terms $n = 51$ to find the mean deviation: \[ \text{MD} = \frac{1300}{51} \approx 25.4901 \approx 25.5 \]
Step 4: Final Answer:
The mean deviation about the mean is $25.5$.
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