List of top Mathematics Questions on Statistics asked in JEE Main

If the mean deviation about the median of the numbers \[ k,\,2k,\,3k,\,\ldots,\,1000k \] is \(500\), then \(k^2\) is equal to
The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation $\alpha$ in this data is replaced by $\beta$, then the mean and variance become $10.1$ and $1.99$, respectively. Then $\alpha+\beta$ equals
Let \( \alpha \) and \( \beta \) respectively be the maximum and the minimum values of the function \( f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right) \). Then \( \alpha + 2\beta \) is equal to :
The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are \( 2, 3, 5, 10, 11, 13, 15, 21 \), then the mean deviation about the median of all the 10 observations is:
If the mean and variance of observations \( x, y, 12, 14, 4, 10, 2, 8 \) and 16 respectively where \( x>y \), then the value of \( 3x - y \) is
Mean deviation about median for \( k, 2k, 3k, \ldots, 1000k \) is 500, then the value of \( k^2 \) is:
If dataset $A=\{1,2,3,\ldots,19\}$ and dataset $B=\{ax+b;\,x\in A\}$. If mean of $B$ is $30$ and variance of $B$ is $750$, then sum of possible values of $b$ is
Consider the 10 observations \(2, 3, 5, 10, 11, 13, 15, 21, a\) and \(b\) such that the mean of observations is \(9\) and variance is \(34.2\). Then the mean deviation about median is:
The mean and variance of the observations x, y, 5, 7, 9, 11, 13, 15 are 10 and 20 respectively. If x $>$ y, then the value of x - y is:
Variates are given as \( -10, -7, -1, x, y, 2, 9, 16 \). If the mean \( (\mu) = \frac{7}{2} \) and variance = \( \frac{293}{4} \), find the mean of \( (1 + x + y), x, y, |y - x| \):
If the mean and variance of observations \( x, y, 12, 14, 4, 10, 2 \) is \(8\) and \(16\) respectively, where \( x>y \), then the value of \( 3x - y \) is:
Let the mean and variance of 10 numbers be $10$ and $2$ respectively. If one number $\alpha$ is replaced by another number $\beta$, then the new mean and variance are $10.1$ and $1.99$ respectively. Find $(\alpha+\beta)$.
If dataset $A=\{1,2,3,\ldots,19\}$ and dataset $B=\{ax+b;\,x\in A\}$. If mean of $B$ is $30$ and variance of $B$ is $750$, then sum of possible values of $b$ is
The following table gives a frequency distribution:
The mean and variance of the above data are \( \mu \) and 19 respectively, where \( \mu \) is an integer. Find the value of \( (\lambda + \mu) \):
Marks obtained by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18, then the total number of students is:

Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is

The frequency distribution of the age of students in a class of 40 students is given below:
\(Age\)151617181920
No. of Students58512xy

If the mean deviation about the median is 1.25, then \(4x + 5y\) is equal to:
Let M denote the median of the following frequency distribution.

\(x_i\)

\(f_i\)  

0 - 42
4 - 84
8 - 127
12 - 168
16 - 206
Then 20M is equal to:
For the following data table

\(x_i\)

\(f_i\)  

0 - 42
4 - 84
8 - 127
12 - 168
16 - 206

Find the value of 20M (where M is median of the data)

Let \( a_1, a_2, \ldots, a_{10} \) be 10 observations such that   \(\sum_{k=1}^{10} a_k = 50 \quad \text{and} \quad \sum_{1 \leq k < j \leq 10} a_k \cdot a_j = 1100\). Then the standard deviation of \( a_1, a_2, \ldots, a_{10} \) is equal to:
Consider 10 observations $x_1, x_2, \ldots, x_{10}$ such that \[\sum_{i=1}^{10} (x_i - \alpha) = 2 \quad \text{and} \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 40,\]where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$, respectively. The value of $\frac{\beta}{\alpha}$ is equal to:
Let the mean and the variance of 6 observation a,b, 68, 44, 48, 60 be 55 and 194, respectively if a > b, then a + 3b is
If the variance of the frequency distribution is 160, then the value of \( c \in \mathbb{N} \) is: \[ \begin{array}{|c|c|c|c|c|c|c|} \hline x & c & 2c & 3c & 4c & 5c & 6c \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} \]
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is
Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and \(\frac{205}{7}\)respectively. Then the mean deviation about the mean of these 7 observations is :