Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ...n.$ Then..
Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ...n.$ Then..
- A
$a \le {\rm{Var}}(x) \le b$
- B
${a^2} \le {\rm{Var}}(x) \le {b^2}$
- C
$\frac{{{a^2}}}{4} \le {\rm{Var}}(x)$
- D
${(b - a)^2} \ge {\rm{Var}}(x)$
Similar Questions
Let $x _1, x _2, \ldots \ldots x _{10}$ be ten observations such that $\sum_{i=1}^{10}\left(x_i-2\right)=30, \sum_{i=1}^{10}\left(x_i-\beta\right)^2=98, \beta>2$ and their variance is $\frac{4}{5}$. If $\mu$ and $\sigma^2$ are respectively the mean and the variance of $2\left( x _1-1\right)+4 \beta, 2\left( x _2-1\right)+$ $4 \beta, \ldots . ., 2\left(x_{10}-1\right)+4 \beta$, then $\frac{\beta \mu}{\sigma^2}$ is equal to :
- [JEE MAIN 2025]
The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.
The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.
The variance of the first $n$ natural numbers is
The variance of the first $n$ natural numbers is
Variance of $^{10}C_0$ , $^{10}C_1$ , $^{10}C_2$ ,.... $^{10}C_{10}$ is
Variance of $^{10}C_0$ , $^{10}C_1$ , $^{10}C_2$ ,.... $^{10}C_{10}$ is
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
- [JEE MAIN 2019]