1
JEE Main 2023 (Online) 29th January Evening Shift
Numerical
+4
-1
Change Language

Let $$X=\{11,12,13,....,40,41\}$$ and $$Y=\{61,62,63,....,90,91\}$$ be the two sets of observations. If $$\overline x $$ and $$\overline y $$ are their respective means and $$\sigma^2$$ is the variance of all the observations in $$\mathrm{X\cup Y}$$, then $$\left| {\overline x + \overline y - {\sigma ^2}} \right|$$ is equal to ____________.

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2
JEE Main 2022 (Online) 29th July Morning Shift
Numerical
+4
-1
Change Language

Let the mean and the variance of 20 observations $$x_{1}, x_{2}, \ldots, x_{20}$$ be 15 and 9 , respectively. For $$\alpha \in \mathbf{R}$$, if the mean of $$\left(x_{1}+\alpha\right)^{2},\left(x_{2}+\alpha\right)^{2}, \ldots,\left(x_{20}+\alpha\right)^{2}$$ is 178 , then the square of the maximum value of $$\alpha$$ is equal to ________.

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3
JEE Main 2022 (Online) 28th July Morning Shift
Numerical
+4
-1
Change Language

Let $$x_{1}, x_{2}, x_{3}, \ldots, x_{20}$$ be in geometric progression with $$x_{1}=3$$ and the common ratio $$\frac{1}{2}$$. A new data is constructed replacing each $$x_{i}$$ by $$\left(x_{i}-i\right)^{2}$$. If $$\bar{x}$$ is the mean of new data, then the greatest integer less than or equal to $$\bar{x}$$ is ____________.

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4
JEE Main 2022 (Online) 27th July Morning Shift
Numerical
+4
-1
Change Language

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _____________.

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