Question

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

• A. 24
• B. 25
• C. 25.5
• D. 26

Hint:

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: $\frac{(51^2 - 1) \times 2}{4 \times 51} = \frac{2600 \times 2}{204} = \frac{5200}{204} \approx 25.49$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The mean deviation about the mean is the average of the absolute differences between each data point and the mean of the data set. For a data set $\{x_1, x_2, \dots, x_n\}$, it is given by: \[ MD(\bar{x}) = \frac{\sum |x_i - \bar{x}|}{n} \]

Step 2: Identifying the Sequence and Calculating the Mean:
The given data set is an arithmetic progression (AP) of odd numbers: $1, 3, 5, \dots, 101$.
• First term ($a$) = 1
• Common difference ($d$) = 2
• Last term ($l$) = 101 To find the number of terms ($n$): \[ 101 = 1 + (n - 1)2 \implies 100 = 2(n - 1) \implies n = 51 \] Since the distribution is symmetric, the mean ($\bar{x}$) is the middle term (the $26^{th}$ term): \[ \bar{x} = 1 + (26 - 1)2 = 1 + 50 = 51 \]

Step 3: Calculating the Sum of Absolute Deviations:
We need to calculate $\sum_{i=1}^{51} |x_i - 51|$. The deviations are: $|1-51|, |3-51|, \dots, |49-51|, |51-51|, |53-51|, \dots, |101-51|$ This results in the sequence: $50, 48, \dots, 2, 0, 2, \dots, 48, 50$. Sum = $2 \times (2 + 4 + 6 + \dots + 50)$ Using the AP sum formula for 25 terms ($a=2, l=50, n=25$): \[ \text{Sum} = 2 \times \left[ \frac{25}{2}(2 + 50) \right] = 25 \times 52 = 1300 \]

Step 4: Final Answer
The mean deviation is: \[ MD(\bar{x}) = \frac{1300}{51} \approx 25.4901 \] Rounding to one decimal place as per the options, we get 25.5.