The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is

  • [JEE MAIN 2020]
  • A

    $3.99$

  • B

    $3.98$

  • C

    $4.02$

  • D

    $4.01$

Similar Questions

The mean and standard deviation of $20$ observations were calculated as $10$ and $2.5$ respectively. It was found that by mistake one data value was taken as $25$ instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :

  • [JEE MAIN 2021]

Determine the mean and standard deviation for the following distribution:

$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}$

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80.$ Then which one of the following gives possible values of $a$ and $b$ $?$ 

  • [AIEEE 2008]

The mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively. On rechecking, it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

If it is replaced by $12$

If the mean and variance of the frequency distribution

$x_i$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ $16$
$f_i$ $4$ $4$ $\alpha$ $15$ $8$ $\beta$ $4$ $5$

are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.

  • [JEE MAIN 2023]