The outcome of each of $30$ items was observed; $10$ items gave an outcome $\frac{1}{2} - d$ each, $10$ items gave outcome $\frac {1}{2}$ each and the remaining $10$ items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac {4}{3}$ then $\left| d \right|$ equals

  • [JEE MAIN 2019]
  • A

    $\frac {2}{3}$

  • B

    $2$

  • C

    $\frac {\sqrt 5}{2}$

  • D

    $\sqrt 2$

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]

The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:

  • [JEE MAIN 2021]

Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a _1, a _2, a _3, \ldots ., a _{100}$ is $25$. Then $S$ is

  • [JEE MAIN 2023]

The diameters of circles (in mm) drawn in a design are given below:

Diameters $33-36$ $37-40$ $41-44$ $45-48$ $49-52$
No. of circles $15$ $17$ $21$ $22$ $25$

Calculate the standard deviation and mean diameter of the circles.

[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]

The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be