If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is

  • [JEE MAIN 2019]
  • A

    $400$

  • B

    $380$

  • C

    $525$

  • D

    $480$

Similar Questions

For a statistical data $x _1, x _2, \ldots, x _{10}$ of $10$ values, a student obtained the mean as $5.5$ and $\sum_{i=1}^{10} x _{ i }^2=371$. He later found that he had noted two values in the data incorrectly as $4$ and $5$ , instead of the correct values $6$ and $8$ , respectively. The variance of the corrected data is

  • [JEE MAIN 2025]

Let $X _{1}, X _{2}, \ldots, X _{18}$ be eighteen observations such that $\sum_{ i =1}^{18}\left( X _{ i }-\alpha\right)=36 \quad$ and $\sum_{i=1}^{18}\left(X_{i}-\beta\right)^{2}=90,$ where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1,$ then the value of $|\alpha-\beta|$ is ...... .

  • [JEE MAIN 2021]

If the mean of the frequency distribution

Class: $0-10$ $10-20$ $20-30$ $30-40$ $40-50$
Frequency $2$ $3$ $x$ $5$ $4$

is $28$ , then its variance is $........$.

  • [JEE MAIN 2023]

Find the mean and variance for the data $6,7,10,12,13,4,8,12$

If $\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $ and $\sum\limits_{i = 1}^{18} {({x_i} - 8)^2 = 45} $ then the standard deviation of $x_1, x_2, ...... x_{18}$ is :-