Statistics JEE Main previous year questions with solutions

5 solved JEE Main questions on Statistics, free to read — no sign-in needed. The full chapter has 164 questions; sign in to attempt the remaining 159 in the exam simulator.

1. Q1JEE Main 2026 (06 Apr, Shift 1)Mean
   <p>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:</p>
   1. A.<p>$2:1$</p>
   2. B.$3:1$
   3. C.$3:2$
   4. D.<p>$4:1$</p>
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   Answer: (B)

   <p>Given $\sum_{i=1}^{20}(x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20}(x_i - 5)^2 = 100$.</p> <p>Expanding the first equation:&nbsp;$\sum_{i=1}^{20} (x_i^2 + 10x_i + 25) = 2500$ $\sum_{i=1}^{20} x_i^2 + 10 \sum_{i=1}^{20} x_i + 500 = 2500$ $\sum_{i=1}^{20} x_i^2 + 10 \sum_{i=1}^{20} x_i = 2000$ --- (1)</p> <p>Expanding the second equation:&nbsp;$\sum_{i=1}^{20} (x_i^2 - 10x_i + 25) = 100$ $\sum_{i=1}^{20} x_i^2 - 10 \sum_{i=1}^{20} x_i + 500 = 100$ $\sum_{i=1}^{20} x_i^2 - 10 \sum_{i=1}^{20} x_i = -400$ --- (2)</p> <p>&nbsp;Subtracting (2) from (1): $20 \sum_{i=1}^{20} x_i = 2400 \Rightarrow \sum_{i=1}^{20} x_i = 120$ Adding (1) and (2): $2 \sum_{i=1}^{20} x_i^2 = 1600 \Rightarrow \sum_{i=1}^{20} x_i^2 = 800$ Mean ($\mu$) is given by: $\mu = \dfrac{\sum_{i=1}^{20} x_i}{20} = \dfrac{120}{20} = 6$</p> <p>&nbsp;Variance ($\sigma^2$) is given by: $\sigma^2 = \dfrac{\sum_{i=1}^{20} x_i^2}{20} - \mu^2 = \dfrac{800}{20} - 6^2 = 40 - 36 = 4$ Standard deviation ($\sigma$) is $\sqrt{4} = 2$.</p> <p>The ratio of mean to standard deviation is:&nbsp;$\dfrac{\mu}{\sigma} = \dfrac{6}{2} = \dfrac{3}{1}$</p> <p>Answer:&nbsp;$3:1$</p>
2. Q2JEE Main 2025 (07 Apr, Shift 1)Measures of Dispersion
   <p>The mean and standard deviation of 100 observations are 40 and 5.1 , respectively, By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are $\mu$ and $\sigma$ respectively, then $10(\mu+\sigma)$ is equal to</p>
   1. A.445
   2. B.451
   3. C.447
   4. D.<p>449</p>
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   Answer: (D)

   <p>$\begin{aligned} & \text { Actual means }=\mu=\frac{100(40)-50+40}{100} \\ & \mu=40-\frac{1}{10}=39.9 \end{aligned}$ Incorrect variance $\begin{aligned} & (5.1)^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{100}-(\overline{\mathrm{x}})^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=100 \times\left(40^2\right)+100(5.1)^2 \\ & \sum \mathrm{x}_{\mathrm{i}}^2=16 \times 10^4+(5.1)^2 \times 100=162601 \\ & \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2-50^2+40^2}{100}-(\mu)^2 \\ & \sigma^2=1617.01-(39.9)^2=25 \\ & \sigma=5 \\ & 10(\mu+\sigma)=10(39.9+5) \\ & =10 \times 44.9=449\end{aligned}$</p>
3. Q3JEE Main 2024 (06 Apr, Shift 1)Mean Square Deviation
   <p>The mean and standard deviation of 20 observations are found to be 10 and 2 . respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is</p>
   1. A.<p>1.8</p>
   2. B.1.94
   3. C.<p>$\sqrt{3.96}$</p>
   4. D.<p>$\sqrt{3.86}$</p>
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   Answer: (C)

   <p>$\begin{aligned} & \text { Mean }(\overline{\mathrm{x}})=10 \\ & \Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}=10 \\ & \Sigma \mathrm{x}_{\mathrm{i}}=10 \times 20=200 \end{aligned}$ If 8 is replaced by 12, then $\Sigma x_i=200-8+12=204$ $\therefore$ Correct mean $(\overline{\mathrm{x}})=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}$ $=\frac{204}{20}=10.2$ $\because$ Standard deviation $=2$ $\therefore$ Variance $=(\text { S.D. })^2=2^2=4$ $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{20}\right)^2=4$ $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}-(10)^2=4$ $\Rightarrow \frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{20}=104$ $\Rightarrow \Sigma x_i^2=2080$ Now, replaced ' 8 ' observations by ' 12 ' Then, $\Sigma \mathrm{x}_{\mathrm{i}}^2=2080-8^2+12^2=2160$ $\therefore$ Variance of removing observations $\begin{aligned} & \Rightarrow \frac{\Sigma x_i^2}{20}-\left(\frac{\Sigma x_i}{20}\right)^2 \\ & \Rightarrow \frac{2160}{20}-(10.2)^2 \\ & \Rightarrow 108-104.04 \\ & \Rightarrow 3.96 \end{aligned}$ Correct standard deviation $=\sqrt{3.96}$</p>
4. Q4JEE Main 2021 (25 Jul, Shift 1)Median
   <p>Consider the following frequency distribution : class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ Frequency $\alpha$ $110$ $54$ $30$ $\beta$ If the sum of all frequencies is $584$ and median is $45$, then $|\alpha -\beta |$ is equal to .</p>
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   Answer: 164

   <p>$∵$ Sum of frequencies $=584$ $\alpha +110+54+30+\beta =584$ $\Rightarrow \alpha +\beta =390...\left(1\right)$ Now, Median is at $\dfrac{584}{2}={292}^{th}$ term $∵$ Median $=45$ (lies in class $40-50$) $\Rightarrow \alpha +110+54+15=292$ $\Rightarrow \alpha =113...\left(2\right)$ On solving, equations $\left(1\right)$ and$\left(2\right)$, we get $\Rightarrow \alpha =113,\beta =390-113=277$ $\Rightarrow |\alpha -\beta |=164$</p>
5. Q5JEE Main 2013 (23 Apr)Relation Between Mean, Median and Mode
   If the median and the range of four numbers $\{x, y, 2 x+y, x-y\}$, where $0 \lt y \lt x \lt 2 y$, are 10 and 28 respectively, then the mean of the numbers is :
   1. A.18
   2. B.10
   3. C.5
   4. D.14
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   Answer: (D)

   Since $0 \lt y \lt x \lt 2 y$ $$\begin{aligned} & \therefore \quad y>\frac{x}{2} \Rightarrow x-y < \frac{x}{2} \\ & \therefore x-y < y < x < 2 x+y \end{aligned}$$ Hence median $=\frac{y+x}{2}=10$ $\Rightarrow x+y=20$ And range $=(2 x+y)-(x-y)=x+2 y$ But range $=28$ $$\therefore x+2 y=28$$ From equations (i) and (ii), $$\begin{aligned} & x=12, y=8 \\ & \therefore \text { Mean } \\ & =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4} \end{aligned}$$ $$=x+\frac{y}{4}=12+\frac{8}{4}=14$$