Let $n \geq 3$. A list of numbers $x_1, x, \ldots, x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers $y_1, y_2, \ldots, y_n$ is made as follows $y_1=\frac{x_1+x_2}{2}, y_2=\frac{x_1+x_2}{2}$ and $y_j=x_j$ for $j=3,4, \ldots, n$.
The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Then, which of the following is necessarily true?
$\mu=\hat{\mu}$ and $\sigma \leq \hat{\sigma}$
$\mu=\hat{\mu}$ and $\sigma \geq \hat{\sigma}$
$\sigma=\hat{\sigma}$
$\mu \neq \hat{\mu}$
If $\sum_{i=1}^{5}(x_i-10)=5$ and $\sum_{i=1}^{5}(x_i-10)^2=5$ then standard deviation of observations $2x_1 + 7, 2x_2 + 7, 2x_3 + 7, 2x_4 + 7$ and $2x_5 + 7$ is equal to-
Let in a series of $2 n$ observations, half of them are equal to $a$ and remaining half are equal to $-a.$ Also by adding a constant $b$ in each of these observations, the mean and standard deviation of new set become $5$ and $20 ,$ respectively. Then the value of $a^{2}+b^{2}$ is equal to ....... .
Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2 \ldots$, $x _7$ is $4$ and the mean is $\overline{ x }$, then $\overline{ x }+ x _6$ is equal to:
Find the mean and variance of the frequency distribution given below:
$\begin{array}{|l|l|l|l|l|} \hline x & 1 \leq x<3 & 3 \leq x<5 & 5 \leq x<7 & 7 \leq x<10 \\ \hline f & 6 & 4 & 5 & 1 \\ \hline \end{array}$