Let x_{i} (1 leq i leq 10) be ten observation | vedclass

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(C) The variance of a set of observations $x_{i}$ is independent of the shift $p$. Let $y_{i} = x_{i} - p$.Given $\sum_{i=1}^{10} y_{i} = 3$ and $\sum

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$\frac{9}{10}$

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(C) The variance of a set of observations $x_{i}$ is independent of the shift $p$. Let $y_{i} = x_{i} - p$.Given $\sum_{i=1}^{10} y_{i} = 3$ and $\sum_{i=1}^{10} y_{i}^{2} = 9$.The variance $\sigma^{2}$ is given by the formula:$\sigma^{2} = \frac{\sum y_{i}^{2}}{n} - \left( \frac{\sum y_{i}}{n} \right)^{2}$Substituting the given values with $n = 10$:$\sigma^{2} = \frac{9}{10} - \left( \frac{3}{10} \right)^{2}$$\sigma^{2} = 0.9 - 0.09 = 0.81$Standard deviation $\sigma = \sqrt{0.81} = 0.9 = \frac{9}{10}$.

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $X$. If $\sum_{i=1}^{10} (x_{i} - p) = 3$ and $\sum_{i=1}^{10} (x_{i} - p)^{2} = 9$,where $0 \neq p \in R$,then the standard deviation of these observations is:

Similar Questions

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{4}$.
Statement-$2$: The sum of the first $n$ natural numbers is $\frac{n(n + 1)}{2}$ and the sum of the squares of the first $n$ natural numbers is $\frac{n(n + 1)(2n + 1)}{6}$.

If the variance of the distribution is $45.8$,then find the variance of the distribution given below:

$x_i$ $4$ $8$ $11$ $17$ $20$ $24$ $32$

$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

$y_i$ $10$ $18$ $24$ $36$ $42$ $50$ $66$

$f_i$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

If variables $x$ and $u$ are related by $u = \frac{x - a}{h}$,then the correct relationship between $\sigma_x$ and $\sigma_u$ is:

The variance of the first $50$ even natural numbers is

The mean of a distribution is $4$. If its coefficient of variation is $58\%$,what is the standard deviation of the distribution?

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