Mathematics · Statistics

JEE Main 2025 — 4 April, Evening Shift — Question 33

Let the mean and the standard deviation of the observation 2,3,3,4,5,7,a,b2,3,3,4,5,7, a, b be 4 and 2\sqrt{2} respectively. Then

the mean deviation about the mode of these observations is :

  1. Option A:

    34\frac{3}{4}

  2. Option B:

    1

    Correct
  3. Option C:

    12\frac{1}{2}

  4. Option D:

    2

Answer: B

Step-by-step solution

2+3+3+4+5+7+a+b8=4\frac{2+3+3+4+5+7+a+b}{8}=4

⇒a+b=8\Rightarrow a+b=8

(2)2=22+32+32+42+52+72+a2+b28−16(\sqrt{2})^{2}=\frac{2^{2}+3^{2}+3^{2}+4^{2}+5^{2}+7^{2}+a^{2}+b^{2}}{8}-16 112+a2+b2=18×8112+a^{2}+b^{2}=18 \times 8

⇒a2+b2=32\Rightarrow a^{2}+b^{2}=32

⇒a=b=4\Rightarrow a=b=4 Now numbers be 2,3,3,4,4,4,5,72,3,3,4,4,4,5,7

Mode =4=4 Mean deviation about mode :

∣2−4∣+∣3−4∣+∣3−4∣+0+0+0+∣5−4∣+∣4−7∣8\frac{|2-4|+|3-4|+|3-4|+0+0+0+|5-4|+|4-7|}{8} =2+1+1+1+38=88=1=\frac{2+1+1+1+3}{8}=\frac{8}{8}=1

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Statistics
Topic
Measures of Central Tendency