Statistics & Linear ProgrammingmediumPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Mean-to-SD Ratio from Sums = 3:1 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
A data consists of 2020 observations x1,x2,…,x20x_1,x_2,\ldots,x_{20}. If ∑i=120(xi+5)2=2500\displaystyle\sum_{i=1}^{20}(x_i+5)^2=2500 and ∑i=120(xi−5)2=100\displaystyle\sum_{i=1}^{20}(x_i-5)^2=100, then the ratio of mean to standard deviation of this data is
A2:12:1
B3:13:1correct
C3:23:2
D4:14:1
Solution
Step 1: Over 2020 observations,
∑(xi+5)2=∑xi2+10∑xi+20(25),∑(xi−5)2=∑xi2−10∑xi+20(25).\sum(x_i+5)^2=\sum x_i^2+10\sum x_i+20(25),\qquad\sum(x_i-5)^2=\sum x_i^2-10\sum x_i+20(25).
 Step 2: Subtract:
20∑xi=2500−100=2400⇒∑xi=120⇒xˉ=12020=6.20\sum x_i=2500-100=2400\Rightarrow\sum x_i=120\Rightarrow\bar x=\dfrac{120}{20}=6.
 Step 3: Add:
2∑xi2+2⋅20(25)=2500+100=2600⇒2∑xi2+1000=2600⇒∑xi2=800.2\sum x_i^2+2\cdot20(25)=2500+100=2600\Rightarrow2\sum x_i^2+1000=2600\Rightarrow\sum x_i^2=800.
 Step 4: σ2=∑xi2n−xˉ2=80020−62=40−36=4⇒σ=2.\sigma^2=\dfrac{\sum x_i^2}{n}-\bar x^2=\dfrac{800}{20}-6^2=40-36=4\Rightarrow\sigma=2. Step 5: ∴xˉσ=62=3\therefore\dfrac{\bar x}{\sigma}=\dfrac{6}{2}=3, i.e. mean : \,:\,SD =3:1=3:1. Correct answer: (2)
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