A data consists of 20 observations x 1 , x 2 , ldots . ., x 20 . If sum i=1 20 (x i +5 ) 2 =2500 and sum i=1 20 (x i -5 ) 2 =100 , then the ratio of mean to standard deviation of this data is:

Mathematics · Statistics

JEE Main 2026 — 6 April, Morning Shift — Question 29

A data consists of 20 observations $x_{1}, x_{2}, \ldots . ., x_{20}$ . If $\sum_{i=1}^{20}\left(x_{i}+5\right)^{2}=2500$ and $\sum_{i=1}^{20}\left(x_{i}-5\right)^{2}=100$ , then the ratio of mean to standard deviation of this data is:

Option A:
$2: 1$
Option B:
$3: 1$
Correct
Option C:
$3: 2$
Option D:
$4: 1$

Answer: B

Step-by-step solution

$\sum_{\mathrm{i}=1}^{20}\left(\mathrm{x}_{\mathrm{i}}+5\right)^{2}=2500$

\sum_{\mathrm{i}=1}^{20}\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}=100 \end{gathered}$$ on (1) - (2) $\Rightarrow 20 \sum \mathrm{x}_{\mathrm{i}}=2400 \Rightarrow \sum \mathrm{x}_{\mathrm{i}}=120$ mean, $\overline{\mathrm{x}}=6$ (1) + (2) $\Rightarrow 2 \sum \mathrm{x}_{\mathrm{i}}^{2}+40\left(5^{2}\right)=2600$ $\sum \mathrm{x}_{\mathrm{i}}^{2}=800$ $\sigma^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{\mathrm{n}}-(\overline{\mathrm{x}})^{2}=\frac{800}{20}-36=4$ $\bar{x}: \sigma=6: 2=3: 1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Statistics
Topic: Measures of Dispersion