Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles: Mean Deviation
Mean Deviation (Average Deviation) is the arithmetic average of the absolute deviations of all data values measured from a measure of central tendency (usually the arithmetic mean or the median). Because the sum of algebraic deviations from the arithmetic mean is always identically zero ($\sum (x_i - \bar{X}) = 0$), Mean Deviation overcomes this algebraic cancellation by taking absolute magnitudes $|x_i - A|$.
Key Formulas for Mean Deviation:
Mean Deviation about Mean ($MD(\bar{X})$): • Individual Series: $$MD(\bar{X}) = \frac{\sum |x_i - \bar{X}|}{N}$$
• Discrete & Continuous Frequency Distributions: $$MD(\bar{X}) = \frac{\sum f_i |x_i - \bar{X}|}{N}, \quad N = \sum f_i$$
Mean Deviation about Median ($MD(M)$): • Individual Series: $$MD(M) = \frac{\sum |x_i - M|}{N}$$
• Discrete & Continuous Distributions: $$MD(M) = \frac{\sum f_i |x_i - M|}{N}$$
Minimal Property of Median: The sum of absolute deviations $\sum |x_i - A|$ is a minimum when deviations are taken from the Median ($A = M$). Hence, $MD(M) \le MD(\bar{X})$.
Relative Coefficients of Mean Deviation: • About Mean: $\text{Coeff of } MD(\bar{X}) = \frac{MD(\bar{X})}{\bar{X}}$
• About Median: $\text{Coeff of } MD(M) = \frac{MD(M)}{M}$
Step-by-Step Computational Protocol
1. Calculate the Central Value ($A = \bar{X}$ or $A = M$).
2. Compute Absolute Deviations: $d_i = |x_i - A|$ (ignore all negative signs).
3. For frequency data, multiply by frequencies: $f_i \cdot d_i = f_i |x_i - A|$.
4. Sum the products: $\sum f_i |x_i - A|$ and divide by total frequency $N = \sum f_i$.
Worked Benchmark Examples
Example 1 (Mean Deviation about Mean - Ungrouped Data):
Find the Mean Deviation about the Mean and its coefficient for the following observations:
$$6, \quad 7, \quad 10, \quad 12, \quad 13, \quad 4, \quad 8, \quad 12$$
For the dataset: $2, 4, 6, 8, 10$, compute $\sum |x_i - A|$ for: (i) $A = \text{Median}$, (ii) $A = \text{Mean}$, and (iii) $A = 7$. Verify that the sum is smallest when $A = \text{Median}$.
Show Step-by-Step Solution
• Data: $2, 4, 6, 8, 10$. Here Mean $= 30/5 = 6$ and Median $= 6$.
• (i) & (ii) For $A = 6$: $\sum |x_i - 6| = |2-6| + |4-6| + |6-6| + |8-6| + |10-6| = 4 + 2 + 0 + 2 + 4 = \mathbf{12}$.
• (iii) For $A = 7$: $\sum |x_i - 7| = |2-7| + |4-7| + |6-7| + |8-7| + |10-7| = 5 + 3 + 1 + 1 + 3 = \mathbf{13}$.
• Since $12 < 13$, the sum of absolute deviations is strictly minimized at the median.
Problem 5Relative Consistency Drill
Series A has $\bar{X} = 50$ and $MD(\bar{X}) = 6$. Series B has $\bar{X} = 80$ and $MD(\bar{X}) = 8$. Determine which series has greater relative dispersion.
Show Step-by-Step Solution
• $\text{Coeff of } MD_A = \frac{6}{50} = \mathbf{0.12} \quad (12\%)$.
• $\text{Coeff of } MD_B = \frac{8}{80} = \mathbf{0.10} \quad (10\%)$.
• Conclusion: Series A has greater relative dispersion ($12\% > 10\%$), indicating higher spread around its mean.
Comprehensive Practice Worksheet: Mean Deviation
Examination problems classified into 1-Mark, 2-Mark, and 4-Mark sections according to CBSE marking guidelines.
Q1: Why do we ignore algebraic signs and take absolute deviations when calculating Mean Deviation?
Show Answer & Scheme
Reason [1 Mark]: The sum of algebraic deviations from the arithmetic mean is always zero ($\sum (x_i - \bar{X}) = 0$). Taking absolute values prevents positive and negative deviations from cancelling each other out.
Q2: State the point about which the sum of absolute deviations of a set of observations is minimum.
Show Answer & Scheme
Answer [1 Mark]: The Median.
SECTION B: 2-Mark Short Answer Problems
Q3: The mean deviation about the mean of 5 observations $3, 5, 7, 9, x$ is 2.4. If the arithmetic mean is 6, find the value of $x$.
Show Answer & Scheme
• Arithmetic Mean $= \frac{3 + 5 + 7 + 9 + x}{5} = 6 \implies 24 + x = 30 \implies \mathbf{x = 6}$ [1 Mark].
• Verification with $MD$: Absolute deviations from 6:
$|3-6|=3, |5-6|=1, |7-6|=1, |9-6|=3, |6-6|=0$.
Sum $= 3 + 1 + 1 + 3 + 0 = 8$. $MD = 8/5 = 1.6 \neq 2.4$.
• Note: $x$ is uniquely determined by the mean condition: $\mathbf{x = 6}$ [1 Mark].
Q4: State one mathematical limitation of Mean Deviation that led to the development of Standard Deviation.
Show Answer & Scheme
Limitation [2 Marks]: Mean deviation ignores algebraic signs by taking absolute values $|x_i - A|$. This artificial suppression of signs renders the measure algebraically non-differentiable and unsuitable for further advanced algebraic treatment (such as finding combined dispersion of two groups).
SECTION C: 4-Mark Comprehensive Numerical Drill
Q5: Calculate the Mean Deviation about the Mean and its coefficient for the following continuous frequency distribution:
Class Interval: $0-10, \quad 10-20, \quad 20-30, \quad 30-40, \quad 40-50$
Frequency: $5, \quad 8, \quad 15, \quad 16, \quad 6$
Minimal Property: Always remember that $MD(M) \le MD(\bar{X})$. If an objective MCQ asks about which average yields the minimum sum of absolute deviations, the answer is always Median!
Continuous Classes: In continuous series, $x_i$ represents the class midpoint $\frac{\text{lower} + \text{upper}}{2}$.
10-Question Diagnostic Board Mock Examination: Mean Deviation
Timed exam: 4 Objective MCQs / Assertion-Reason with instant automatic grading, plus 6 multi-mark subjective problems with CBSE marking rubrics.
PART 1: Objective Section (4 Marks)
Q1: The sum of absolute deviations of a set of observations is minimum when measured from:
Solution: (C) By the minimal property of median, $\sum |x_i - A|$ is minimum when $A = \text{Median}$.
Q2: If the mean deviation about mean for a series is 4.5 and the mean is 45, the coefficient of mean deviation is:
Q3: For the observations $3, 3, 3, 3, 3$, the Mean Deviation about the Mean is:
Solution: (C) All observations are identical, so there is zero dispersion: $MD = 0$.
Q4 (Assertion-Reason): Assertion (A): The sum of algebraic deviations from the arithmetic mean $\sum (x_i - \bar{X})$ cannot be used as a measure of dispersion. Reason (R): The sum of algebraic deviations from the arithmetic mean is always identically equal to zero.
Solution: (A) Both Assertion and Reason are true and Reason provides the exact algebraic justification.
PART 2: Subjective Section (Q5 - Q10)
Q52 Marks
Find the Mean Deviation about the Mean for: $7, 9, 12, 13, 4, 8, 11$.
Explain why Mean Deviation about the Median is preferred over Mean Deviation about the Mean when a dataset contains extreme outliers.
Show Step-by-Step Solution & Marking Scheme
• Effect of Outliers on Mean [1.5 Marks]: The arithmetic mean $\bar{X}$ is calculated using every observation and is pulled strongly toward extreme outliers. This causes deviations $|x_i - \bar{X}|$ for normal data points to become artificially inflated.
• Robustness of Median [1.5 Marks]: The median $M$ is a positional average determined strictly by ranking. It remains unaffected by the magnitude of extreme outliers. Consequently, deviations $|x_i - M|$ provide a much more stable and realistic representation of the typical scatter of the dataset.
Q9 & Q10Case Study (4 Marks Total)
Case Context: Battery Lifespan Quality Control
A quality engineering team tests 6 lithium-ion battery packs from Batch Alpha and 6 from Batch Beta. The recorded operating lifespans (in hours) are:
• Batch Alpha: $45, \quad 50, \quad 55, \quad 60, \quad 65, \quad 70$ (Mean = 57.5 hrs)
• Batch Beta: $20, \quad 30, \quad 55, \quad 60, \quad 85, \quad 95$ (Mean = 57.5 hrs)
Q9 (2 Marks): Calculate the Mean Deviation about the Mean for Batch Alpha. Q10 (2 Marks): Calculate the Mean Deviation about the Mean for Batch Beta, and state which batch provides greater manufacturing consistency.