Statistics JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Statistics, free to read — no sign-in needed. The full chapter has 4 questions; sign in to attempt the remaining 1 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Consider a data consisting of $10$ observations $x_1, x_2, \ldots, x_{10}$, whose mean is $5$ and variance is $7$. If the mean and the variance of the first $8$ observations $x_1, x_2, \ldots, x_8$ are $4$ and $3.5$, respectively, and $x_9 \lt x_{10}$, then the value of $3x_9 + 2x_{10}$ is ___________.
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   Answer: 44

   Given $\dfrac{\sum_{i=1}^{10} x_i}{10} = 5 \Rightarrow \sum_{i=1}^{10} x_i = 50$ $\dfrac{\sum_{i=1}^{10} x_i^2}{10} - 5^2 = 7 \Rightarrow \sum_{i=1}^{10} x_i^2 = 320$ For the first $8$ observations: $\dfrac{\sum_{i=1}^{8} x_i}{8} = 4 \Rightarrow \sum_{i=1}^{8} x_i = 32$ $\dfrac{\sum_{i=1}^{8} x_i^2}{8} - 4^2 = 3.5 \Rightarrow \sum_{i=1}^{8} x_i^2 = 8(3.5 + 16) = 156$ Subtracting the sums, we get: $x_9 + x_{10} = 50 - 32 = 18$ $x_9^2 + x_{10}^2 = 320 - 156 = 164$ Using $(x_9 + x_{10})^2 = x_9^2 + x_{10}^2 + 2x_9 x_{10}$: $18^2 = 164 + 2x_9 x_{10} \Rightarrow 2x_9 x_{10} = 324 - 164 = 160 \Rightarrow x_9 x_{10} = 80$ The numbers $x_9$ and $x_{10}$ are roots of the equation $t^2 - 18t + 80 = 0$. $(t - 8)(t - 10) = 0 \Rightarrow t = 8, 10$ Since $x_9 \lt x_{10}$, we have $x_9 = 8$ and $x_{10} = 10$. Therefore, $3x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44$. Answer: $44$
2. Q2JEE Advanced Adv 2024 (Paper 1)
   Let $X$ be a random variable, and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x)), x=0,1,2,3,4$, lie on a fixed straight line in the $x y$-plane, and $P(X=x)=0$ for all $x \in \mathbb{R}-\{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$, and the variance of $X$ is $\alpha$, then the value of $24 \alpha$ is ___
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   Answer: 42

   Let equation of line is y = mx + c $\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{x} & 0 & \mathrm{l} & 2 & 3 & 4 & \mathrm{R}-\{0,1,2,3,4\} \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{c} & \mathrm{m}+\mathrm{c} & 2 \mathrm{~m}+\mathrm{c} & 3 \mathrm{~m}+\mathrm{c} & 4 \mathrm{~m}+\mathrm{c} & 0 \\ \hline \end{array}$ $\sum_{x=0}^4(m x+c)=1 \Rightarrow 10 m+5 c=1 \Rightarrow 2 m+c=\frac{1}{5}...(1)$ $\begin{aligned} & \text { mean }=\sum x_i P_i=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \cdot \mathrm{x}_{\mathrm{i}}=30 \mathrm{~m}+10 \mathrm{c}=\frac{5}{2} \\ & \therefore 3 \mathrm{~m}+\mathrm{c}=\frac{1}{4} \ldots(2)\end{aligned}$ $\begin{aligned} & \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \mathrm{x}_1^2 \\ & =\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}^3+\mathrm{cx}_{\mathrm{i}}^2\right) \Rightarrow 100 \mathrm{~m}+30 \mathrm{c} \text { (Now putting } \mathrm{m} \text { and } \mathrm{c} \text { ) } \\ & \Rightarrow \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=5+3=8 \\ & \text { Variance }=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4} \\ & \therefore 24 \alpha=42\end{aligned}$
3. Q3JEE Advanced Adv 2023 (Paper 1)
   Consider the given data with frequency distribution $\begin{matrix}{x}_{i} & 3 & 8 & 11 & 10 & 5 & 4 \\ {f}_{i} & 5 & 2 & 3 & 2 & 4 & 4\end{matrix}$ Match each entry in List-I to the correct entries in List-II. List-I List-II $\left(P\right)$ The mean of the above data is $\left(1\right)$ $2.5$ $\left(Q\right)$ The median of the above data is $\left(2\right)$ $5$ $\left(R\right)$ The mean deviation about the mean of the above data is $\left(3\right)$ $6$ $\left(S\right)$ The mean deviation about the median of the above data is $\left(4\right)$ $2.7$ $\left(5\right)$ $2.4$ The correct option is
   1. A.$\left(P\right)\rightarrow \left(3\right)\left(Q\right)\rightarrow \left(2\right)\left(R\right)\rightarrow \left(4\right)\left(S\right)\rightarrow \left(5\right)$
   2. B.$\left(P\right)\rightarrow \left(3\right)\left(Q\right)\rightarrow \left(2\right)\left(R\right)\rightarrow \left(1\right)\left(S\right)\rightarrow \left(5\right)$
   3. C.$\left(P\right)\rightarrow \left(2\right)\left(Q\right)\rightarrow \left(3\right)\left(R\right)\rightarrow \left(4\right)\left(S\right)\rightarrow \left(1\right)$
   4. D.$\left(P\right)\rightarrow \left(3\right)\left(Q\right)\rightarrow \left(3\right)\left(R\right)\rightarrow \left(5\right)\left(S\right)\rightarrow \left(5\right)$
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   Answer: (A)

   Arranging the given data in ascending order we get, $\begin{matrix}{x}_{i} & 3 & 4 & 5 & 8 & 10 & 11 \\ {f}_{i} & 5 & 4 & 4 & 2 & 2 & 3\end{matrix}$ Now finding the mean of the above data we get, Mean $=\dfrac{3\times 5+8\times 2+11\times 3+10\times 2+5\times 4+4\times 4}{5+2+3+2+4+4}$ $=\dfrac{15+16+33+20+20+16}{20}=\dfrac{120}{20}=6$ Now, median $=\dfrac{1}{2}\left({10}^{th}+{11}^{th}observation\right)$ $=\dfrac{1}{2}\left(5+5\right)=5$ Hence, Mean deviation about mean will be, $=\dfrac{3\times 5+2\times 4+1\times 4+2\times 2+4\times 2+5\times 3}{20}$ $=\dfrac{54}{20}=2.7$ And Mean deviation about median $=\dfrac{2\times 5+1\times 4+0+3\times 2+5\times 2+6\times 3}{20}$ $=\dfrac{4.8}{20}=2.4$ $P\rightarrow 3;Q\rightarrow 2;R\rightarrow 4;S\rightarrow 5$ $∴$ Option A is correct.