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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Statistics

For 10 observations x_1, x_2, , x_ 10 , if ∑_ i=1 ^ 10...

For 1010 observations x1,x2,…,x10x_1, x_2, \ldots, x_{10}, if ∑i=110(xi+2)2=180\sum_{i=1}^{10}(x_i+2)^2=180 and ∑i=110(xi−1)2=90\sum_{i=1}^{10}(x_i-1)^2=90, then their standard deviation is:

Options

  1. A22
  2. B3\sqrt{3}
  3. C222\sqrt{2}
  4. D33

Correct answer

D. 33

Step-by-step solution

Given ∑i=110(xi+2)2=180\sum_{i=1}^{10}(x_i+2)^2=180

⇒∑i=110xi2+4∑i=110xi+40=180\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i + 40 = 180

⇒∑i=110xi2+4∑i=110xi=140\Rightarrow \sum_{i=1}^{10} x_i^2 + 4\sum_{i=1}^{10} x_i = 140

Also given ∑i=110(xi−1)2=90\sum_{i=1}^{10}(x_i-1)^2=90

⇒∑i=110xi2−2∑i=110xi+10=90\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i + 10 = 90

⇒∑i=110xi2−2∑i=110xi=80\Rightarrow \sum_{i=1}^{10} x_i^2 - 2\sum_{i=1}^{10} x_i = 80

Subtracting the second equation from the first:

6∑i=110xi=60⇒∑i=110xi=106\sum_{i=1}^{10} x_i = 60 \Rightarrow \sum_{i=1}^{10} x_i = 10

Substituting this value in the second equation:

∑i=110xi2−2(10)=80⇒∑i=110xi2=100\sum_{i=1}^{10} x_i^2 - 2(10) = 80 \Rightarrow \sum_{i=1}^{10} x_i^2 = 100

The standard deviation is given by:

σ=∑xi2n−(∑xin)2\sigma = \sqrt{\dfrac{\sum x_i^2}{n} - \left(\dfrac{\sum x_i}{n}\right)^2}

σ=10010−(1010)2\sigma = \sqrt{\dfrac{100}{10} - \left(\dfrac{10}{10}\right)^2}

σ=10−1=9=3\sigma = \sqrt{10 - 1} = \sqrt{9} = 3

Answer: 33
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