Consider a set of $3 n$ numbers having variance $4.$ In this set, the mean of first $2 n$ numbers is $6$ and the mean of the remaining $n$ numbers is $3.$ A new set is constructed by adding $1$ into each of first $2 n$ numbers, and subtracting $1$ from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9 k$ is equal to .... .

  • [JEE MAIN 2021]
  • A

    $76$

  • B

    $68$

  • C

    $82$

  • D

    $56$

Similar Questions

The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$...................$

  • [JEE MAIN 2024]

Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$

  • [AIEEE 2012]

The means of five observations is $4$ and their variance is $5.2$. If three of these observations are $1, 2$ and $6$, then the other two are

Let $x _1, x _2, \ldots \ldots x _{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2\left( x _1-1\right)+4 \beta, 2\left( x _2-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^2}$ is equal to :

  • [JEE MAIN 2025]

Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is