Mean Deviation of data 3, 10, 10, 4, 7, 18, 5 from mode is

Option c: 4.14. Solution: We are given the dataset: 3, 10, 10, 4, 7, 18, 5 . The number of observations is n = 7 . 1. Identify the Mode: The mode is the observation with the highest frequency. Here, 10 appears twice while all other observations appear once. Thus, Mode = 10 . 2. Calculate absolute deviations from the mode |xi - Mode| : - |3 - 10| = 7 - |10 - 10| = 0 - |10 - 10| = 0 - |4 - 10| = 6 - |7 - 10| = 3 - |18 - 10| = 8 - |5 - 10| = 5 3. Calculate sum of absolute deviations: sum |xi - Mode| = 7 + 0 + 0 + 6 + 3 + 8 + 5 = 29 4. Calculate Mean Deviation about Mode: M.D. = 29/7 ≈ 4.14 Hence, Option C is the correct answer.

Mean Deviation of data $3, 10, 10, 4, 7, 18, 5$ from mode is

The correct answer is Option c: 4.14. We are given the dataset: $3, 10, 10, 4, 7, 18, 5$. The number of observations is $n = 7$. 1. Identify the Mode: The mode is the observation with the highest frequency. Here, $10$ appears twice while all other observations appear once. Thus, $\text{Mode} = 10$. 2. Calculate absolute deviations from the mode $|x_i - \text{Mode}|$: - $|3 - 10| = 7$ - $|10 - 10| = 0$ - $|10 - 10| = 0$ - $|4 - 10| = 6$ - $|7 - 10| = 3$ - $|18 - 10| = 8$ - $|5 - 10| = 5$ 3. Calculate sum of absolute deviations: $$\sum |x_i - \text{Mode}| = 7 + 0 + 0 + 6 + 3 + 8 + 5 = 29$$ 4. Calculate Mean Deviation about Mode: $$\text{M.D.} = \frac{29}{7} \approx 4.14$$ Hence, **Option C** is the correct answer.

Measures of Central Tendency and DispersionMCQPYQ June 22Question 3097 of 473

All Questions

Mean Deviation of data $\displaystyle 3, 10, 10, 4, 7, 18, 5$ from mode is

Options

A4.39

B4.70

C4.14

D5.24

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — 4.14

All Options:

A4.39
B4.70
C4.14
D5.24

Detailed Solution & Explanation

We are given the dataset:

\displaystyle 3, 10, 10, 4, 7, 18, 5

. The number of observations is

\displaystyle n = 7

. 1. Identify the Mode: The mode is the observation with the highest frequency. Here,

\displaystyle 10

appears twice while all other observations appear once. Thus,

\displaystyle \text{Mode} = 10

. 2. Calculate absolute deviations from the mode

\displaystyle |x_i - \text{Mode}|

: -

\displaystyle |3 - 10| = 7

\displaystyle |10 - 10| = 0

\displaystyle |10 - 10| = 0

\displaystyle |4 - 10| = 6

\displaystyle |7 - 10| = 3

\displaystyle |18 - 10| = 8

\displaystyle |5 - 10| = 5

3. Calculate sum of absolute deviations:

\sum |x_i - \text{Mode}| = 7 + 0 + 0 + 6 + 3 + 8 + 5 = 29

4. Calculate Mean Deviation about Mode:

\text{M.D.} = \frac{29}{7} \approx 4.14

Hence, **Option C** is the correct answer.

About This Chapter: Measures of Central Tendency and Dispersion

Paper

Paper 3: Quantitative Aptitude

Weightage

12-15 Marks

Key Topics

Mean, Median, Mode, Range, Mean Deviation, Standard Deviation

The core foundation of Statistics. This chapter covers Mean (Arithmetic, Geometric, Harmonic), Median, Mode, and their properties. It also explores measures of spread like Range, Mean Deviation, Standard Deviation, and Quartile Deviation.

View Official ICAI Syllabus

Exam Strategy Tip

Do not just memorize formulas; ICAI loves asking about the mathematical properties (e.g., 'sum of deviations from the AM is always zero'). You can usually eliminate 2 options just by knowing the properties.

Key Concepts to Understand

Mode

The value that appears most frequently in a data set. A data set can be unimodal (one mode), bimodal (two modes), or multimodal (more than two modes). It is the only measure of central tendency applicable to qualitative (categorical) data.

Related Comparison Tables

Mean vs Median vs Mode: Measures of Central Tendency Fiscal Policy vs Monetary Policy: CA Foundation Economics

Ready to Master Measures of Central Tendency and Dispersion?

Practice all 473 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free

Mean Deviation of data 3,10,10,4,7,18,5\displaystyle 3, 10, 10, 4, 7, 18, 53,10,10,4,7,18,5 from mode is