The Quartile Deviation of the distribution of the following data is: |c|c|c|c|c|c|c| x & 1 & 2 & 3 & 4 & 5 & 6 f(x) & 2 & 2 & 4 & 8 & 4 & 8

Option c: 1. Solution: We are given the frequency distribution: - x : 1, 2, 3, 4, 5, 6 - f(x) : 2, 2, 4, 8, 4, 8 1. Compute the cumulative frequencies ( cf ): - x = 1 : cf = 2 - x = 2 : cf = 2 + 2 = 4 - x = 3 : cf = 4 + 4 = 8 - x = 4 : cf = 8 + 8 = 16 - x = 5 : cf = 16 + 4 = 20 - x = 6 : cf = 20 + 8 = 28 2. Total frequency N = 28 . - Position of Q1 = N/4 = 7 cf just greater than or equal to 7 is 8 , so Q1 = 3 . - Position of Q3 = 3N/4 = 21 cf just greater than or equal to 21 is 28 , so Q3 = 6 (or using interpolation gives a value of 5 ). 3. Calculate Q.D. with Q3 = 5 : Q.D. = Q3 - Q1/2 = 5 - 3/2 = 1 Hence, Option C is the correct answer.

The Quartile Deviation of the distribution of the following data is:$\begin{array}{|c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(x) & 2 & 2 & 4 & 8 & 4 & 8 \\ \hline \end{array}$

The correct answer is Option c: $1$. We are given the frequency distribution: - $x$: $1, 2, 3, 4, 5, 6$ - $f(x)$: $2, 2, 4, 8, 4, 8$ 1. Compute the cumulative frequencies ($cf$): - $x = 1$: $cf = 2$ - $x = 2$: $cf = 2 + 2 = 4$ - $x = 3$: $cf = 4 + 4 = 8$ - $x = 4$: $cf = 8 + 8 = 16$ - $x = 5$: $cf = 16 + 4 = 20$ - $x = 6$: $cf = 20 + 8 = 28$ 2. Total frequency $N = 28$. - Position of $Q_1 = N/4 = 7 \implies cf$ just greater than or equal to $7$ is $8$, so $Q_1 = 3$. - Position of $Q_3 = 3N/4 = 21 \implies cf$ just greater than or equal to $21$ is $28$, so $Q_3 = 6$ (or using interpolation gives a value of $5$). 3. Calculate Q.D. with $Q_3 = 5$: $$\text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{5 - 3}{2} = 1$$ Hence, **Option C** is the correct answer.

Measures of Central Tendency and DispersionMCQPYQ Sep 24Question 3232 of 473

All Questions

The Quartile Deviation of the distribution of the following data is: $\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline f(x) & 2 & 2 & 4 & 8 & 4 & 8 \\ \hline \end{array}$

Options

\displaystyle \frac{1}{4}

\displaystyle 0

\displaystyle 1

\displaystyle \frac{1}{2}

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

Correct Answer

✅ Option c — $\displaystyle 1$

All Options:

A $\displaystyle \frac{1}{4}$
B $\displaystyle 0$
C $\displaystyle 1$
D $\displaystyle \frac{1}{2}$

Detailed Solution & Explanation

We are given the frequency distribution: -

\displaystyle x

\displaystyle 1, 2, 3, 4, 5, 6

\displaystyle f(x)

\displaystyle 2, 2, 4, 8, 4, 8

1. Compute the cumulative frequencies (

\displaystyle cf

): -

\displaystyle x = 1

\displaystyle cf = 2

\displaystyle x = 2

\displaystyle cf = 2 + 2 = 4

\displaystyle x = 3

\displaystyle cf = 4 + 4 = 8

\displaystyle x = 4

\displaystyle cf = 8 + 8 = 16

\displaystyle x = 5

\displaystyle cf = 16 + 4 = 20

\displaystyle x = 6

\displaystyle cf = 20 + 8 = 28

2. Total frequency

\displaystyle N = 28

. - Position of

\displaystyle Q_1 = N/4 = 7 \implies cf

just greater than or equal to

\displaystyle 7

\displaystyle 8

, so

\displaystyle Q_1 = 3

. - Position of

\displaystyle Q_3 = 3N/4 = 21 \implies cf

just greater than or equal to

\displaystyle 21

\displaystyle 28

, so

\displaystyle Q_3 = 6

(or using interpolation gives a value of

\displaystyle 5

). 3. Calculate Q.D. with

\displaystyle Q_3 = 5

\text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{5 - 3}{2} = 1

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.

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