Statistics JEE Main PYQs (2020–2026)

Solved Statistics PYQs

3 recent questions with answers. The full set of 105 is free to practise.

1. JEE Main 2026EasyMean Deviation about Mean and Median
   Q1

   The mean deviation about the mean for the data
   
   $$\begin{array}{ccccccc}{x}_i& 5& 7& 9& 10& 12& 15\\ f_i& 8& 6& 2& 2& 2& 6\\ & & & & & & \end{array}$$
   
   is equal to:

   • A$\frac{40}{13}$
   • B$\frac{42}{13}$
   • C$\frac{44}{13}$
   • D$\frac{46}{13}$

   Show answer & solution

   Correct answer: (C)

   For a discrete frequency distribution, the mean deviation about the mean is
   
   $$\text{M.D. about mean}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i}$$
   
   First, find the mean $\bar{x}$ .
   
   Step 1: Calculate $\sum f_i$ and $\sum f_i{x}_i$
   
   $$\begin{array}{ccc}{x}_i& f_i& f_i{x}_i\\ 5& 8& 40\\ 7& 6& 42\\ 9& 2& 18\\ 10& 2& 20\\ 12& 2& 24\\ 15& 6& 90\\ & & \end{array}$$
   
   Now,
   
   $$\sum f_i=8+6+2+2+2+6=26$$
   
   and
   
   $$\sum f_i{x}_i=40+42+18+20+24+90=234$$
   
   So, the mean is
   
   $$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{234}{26}=9$$
   
   Step 2: Calculate $\sum f_i|x_i-\bar{x}|$
   
   Since $\bar{x}=9$ , we find $|x_i-9|$ :
   
   $$\begin{array}{cccc}{x}_i& f_i& |x_i-9|& f_i|x_i-9|\\ 5& 8& 4& 32\\ 7& 6& 2& 12\\ 9& 2& 0& 0\\ 10& 2& 1& 2\\ 12& 2& 3& 6\\ 15& 6& 6& 36\\ & & & \end{array}$$
   
   Now,
   
   $$\sum f_i|x_i-\bar{x}|=32+12+0+2+6+36=88$$
   
   Step 3: Find mean deviation about the mean
   
   $$\text{M.D. about mean}=\frac{88}{26}=\frac{44}{13}$$
   
   So, the correct answer is
   
   $$\boxed{\frac{44}{13}}$$
   
   Hence, Option C is correct.
2. JEE Main 2026EasyVariance and Standard Deviation of Ungrouped Data
   Q2

   For 10 observations $x_1,x_2,\dots ,x_{10}$ , if ${\sum }_{i=1}^{10}{\left(x_i+2\right)}^2=180$ and ${\sum }_{i=1}^{10}{\left(x_i-1\right)}^2=90$ , then their standard deviation is :

   • A2
   • B$\sqrt{3}$
   • C$2\sqrt{2}$
   • D3

   Show answer & solution

   Correct answer: (D)

   $$\sum _{i=1}^{10}(x_i+2{)}^2=180$$
   
   $$\sum _{i=1}^{10}{x}_i^2+4\sum _{i=1}^{10}{x}_i+\sum _{i=1}^{10}4=180$$
   
   $$\sum _{i=1}^{10}{x}_i^2+4\sum _{i=1}^{10}{x}_i=180-40\dots (1)$$
   
   Also ${\sum }_{i=1}^{10}(x_i-1{)}^2=90$
   
   $$\sum _{i=1}^{10}{x}_i^2-2\sum _{i=1}^{10}{x}_i+\sum _{i=1}^{10}1=90$$
   
   $$\sum _{i=1}^{10}{x}_i^2-2\sum _{i=1}^{10}{x}_i=90-10\dots (2)$$
   
   $$\sum _{i=1}^{10}{x}_i^2+4\sum _{i=1}^{10}{x}_i=140$$
   
   $$\sum _{i=1}^{10}{x}_i^2-2\sum _{i=1}^{10}{x}_i=80$$
   
   $$\sum _{i=1}^{10}{x}_i^2=100\text{and}\sum _{i=1}^{10}{x}_i=10$$
   
   $${\sigma }^2=\frac{{\sum }_{i=1}^{10}{x}_i^2}{N}-{\left(\frac{{\sum }_{i=1}^{10}{x}_i}{N}\right)}^2$$
   
   $$=\frac{100}{10}-{\left(\frac{10}{10}\right)}^2$$
   
   $${\sigma }^2=10-1=9$$
   
   $$\Rightarrow \sigma =3$$
3. JEE Main 2026EasyCorrections of Mean and Variance due to Incorrect Observations
   Q3

   The mean and variance of a data of 10 observations are 10 and 2 , respectively. If an observations $\alpha$ in this data is replaced by $\beta$ , then the mean and variance become 10.1 and 1.99 , respectively. Then $\alpha +\beta$ equals

   • A15
   • B10
   • C5
   • D20

   Show answer & solution

   Correct answer: (D)

   Given, total observation $n=10$
   
   Initial mean ${\bar{x}}_1=10$
   
   $$\Rightarrow \frac{{\sum }_{i=1}^{10}{x}_i}{10}=10$$
   
   Sum of 10 observation ${\sum }_{i=1}^{10}{x}_i=10\times 10=100$
   
   Initial variance ${\sigma }_1^2$ $=2$
   
   $$\begin{array}{rl}& \Rightarrow \frac{{\sum }_{i=1}^{10}{x}_i^2}{n}-{\left({\bar{x}}_1\right)}^2=2\\ \Rightarrow \sum _{i=1}^{10}{x}_i^2=n\left[2+{\left({\bar{x}}_1\right)}^2\right]& \\ =10\left[2+\left(10^2\right)\right]=1020& \end{array}$$
   
   It is given that if $\alpha$ is replace by $\beta$
   
   then, new mean $\left({\bar{x}}_2\right)=10.1$
   
   $$\begin{array}{rl}& {\bar{x}}_2=\frac{100-\alpha +\beta }{10}=10.1\\ & \Rightarrow 100-\alpha +\beta =101\\ & \Rightarrow \beta -\alpha =\mathrm{1......}(1)\end{array}$$
   
   Also new variance ${\sigma }_2^2=1.99$ (given)
   
   $\Rightarrow$ ${\sigma }_2^2=\frac{1020-{\alpha }^2+{\beta }^2}{10}-(10.1{)}^2=1.99$
   
   $\Rightarrow$ $1020-{\alpha }^2+{\beta }^2=10[1.99+102.01]$
   
   $$=10\times 104=1040$$
   
   $$\begin{array}{rl}& \Rightarrow {\beta }^2-{\alpha }^2=20\\ & \\ & \Rightarrow (\beta +\alpha )(\beta -\alpha )=20\end{array}$$
   
   $\Rightarrow$ $(\beta +\alpha )(1)=20$ (substitute $\beta -\alpha =1)$
   
   $$\Rightarrow$$
   
   $$\alpha +\beta =20$$

Topic-wise question count

Where JEE Main has asked from Statistics, most frequent first.

• Variance and Standard Deviation of Ungrouped Data40
• Corrections of Mean and Variance due to Incorrect Observations17
• Mean Deviation about Mean and Median14
• Properties of Variance and SD under Change of Origin and Scale14
• Variance and Standard Deviation of Grouped / Continuous Data12
• Mean, Median and Mode of Grouped and Ungrouped Data8

Year-wise Statistics questions in JEE Main

Number of questions from this chapter in each year's paper(s).