Let the mean and the standard deviation of the observation 2, 3, 3, 4, 5, 7, a, b be 4 and √2 respectively. Then the mean deviation about the mode of these observations is :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q8 – JEE-MAIN 2025

← Back

Home Maths StatisticsQuestion #432

Let the mean and the standard deviation of the observation

2, 3, 3, 4, 5, 7, a, b

4

and

\sqrt{2}

respectively. Then the mean deviation about the mode of these observations is :

Clue 1 of 3: Understand the definitions

The problem requires calculating the mean deviation about the mode, which first necessitates finding the values of $a$ and $b$ using the given mean and standard deviation.

🧠 Full Solution Path

Step 1: Determine the values of

a

and

b

✦ Active

Given $n = 8$ observations: $2, 3, 3, 4, 5, 7, a, b$ . The mean $\bar{x} = 4$ and standard deviation $σ = \sqrt{2}$ . Using the mean formula: $\bar{x} = \frac{\sum x_{i}}{n}$

4 = \frac{2 + 3 + 3 + 4 + 5 + 7 + a + b}{8} \Rightarrow 32 = 24 + a + b \Rightarrow a + b = 8 (1)

Using the variance formula: $σ^{2} = \frac{\sum x_{i}^{2}}{n} - (\bar{x})^{2}$

(\sqrt{2})^{2} = \frac{2^{2} + 3^{2} + 3^{2} + 4^{2} + 5^{2} + 7^{2} + a^{2} + b^{2}}{8} - 4^{2}

This simplifies to:

2 = \frac{4 + 9 + 9 + 16 + 25 + 49 + a^{2} + b^{2}}{8} - 16

Further simplification gives:

2 = \frac{112 + a^{2} + b^{2}}{8} - 16 \Rightarrow 18 = \frac{112 + a^{2} + b^{2}}{8} \Rightarrow 144 = 112 + a^{2} + b^{2} \Rightarrow a^{2} + b^{2} = 32 (2)

From $(1)$ , $b = 8 - a$ . Substitute into $(2)$ :

a^{2} + (8 - a)^{2} = 32 \Rightarrow a^{2} + 64 - 16 a + a^{2} = 32 \Rightarrow 2 a^{2} - 16 a + 32 = 0

Dividing by 2, we get $a^{2} - 8 a + 16 = 0$ , which is $(a - 4)^{2} = 0$ . Thus, $a = 4$ . Substituting $a = 4$ into $a + b = 8$ , we get $4 + b = 8$ , so $b = 4$ .

Step 2: Identify the mode of the observations○ Expand

The complete set of observations is $2, 3, 3, 4, 5, 7, 4, 4$ . Arranging them in ascending order: $2, 3, 3, 4, 4, 4, 5, 7$ . The mode is the value that appears most frequently. In this set, $4$ appears three times, which is more than any other value.

Mode = 4

Step 3: Calculate the mean deviation about the mode○ Expand

The formula for mean deviation about the mode is $M_{d} = \frac{\sum | x_{i} - Mode |}{n}$ . With Mode $= 4$ and $n = 8$ , we calculate the sum of absolute deviations:

\sum | x_{i} - 4 | = | 2 - 4 | + | 3 - 4 | + | 3 - 4 | + | 4 - 4 | + | 4 - 4 | + | 4 - 4 | + | 5 - 4 | + | 7 - 4 |

This sum is:

= 2 + 1 + 1 + 0 + 0 + 0 + 1 + 3 = 8

Therefore, the mean deviation about the mode is:

M_{d} = \frac{8}{8} = 1

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 8 – JEE-MAIN 2025

Clue 1 of 3: Understand the definitions

🧠 Full Solution Path