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 if the wrong item is omitted.

  • A
    $10.1, 2.02$
  • B
    $10.0, 2.00$
  • C
    $9.5, 1.95$
  • D
    $10.5, 2.10$

Explore More

Similar Questions

Let $x_1, x_2, \ldots, x_{11}$ be the observations satisfying $\sum_{i=1}^{11}(x_i-4)=22$ and $\sum_{i=1}^{11}(x_i-4)^2=154$. If the mean and variance of the observations are $\alpha$ and $\beta$,then the quadratic equation having the roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

If both mean and variance of $50$ observations $x_1, x_2, \ldots, x_{50}$ are equal to $16$ and $256$ respectively,then the mean of $(x_1-5)^2, (x_2-5)^2, \ldots, (x_{50}-5)^2$ is

If $x_1, x_2, \ldots, x_n$ are $n$ observations and $\bar{x}$ is their mean. If $\sum_{i=1}^{n}(x_i - \bar{x})^2$ is almost zero,then which of the following statements is true?

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.8$. Then which of the following gives possible values of $a$ and $b$?

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.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo