Let $n \geq 3$. A list of numbers $0 < x_1 < x_2 < \ldots < x_n$ has mean $\mu$ and standard deviation $\sigma$. A new list of numbers is made as follows: $y_1=0, y_2=x_2, \ldots, x_{n-1}$ $=x_n-1, y_n=x_1+x_n$. The mean and the standard deviation of the new list are $\hat{\mu}$ and $\hat{\sigma}$. Which of the following is necessarily true?
$\mu=\hat{\mu}, \sigma \leq \hat{\sigma}$
$\mu=\hat{\mu}, \sigma \geq \hat{\sigma}$
$\sigma=\hat{\sigma}$
$\mu$ may or may not be equal to $\hat{\mu}$
Variance of $^{10}C_0$ , $^{10}C_1$ , $^{10}C_2$ ,.... $^{10}C_{10}$ is
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?
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:
If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to
Mean and standard deviation of 100 items are 50 and $4,$ respectively. Then find the sum of all the item and the sum of the squares of the items.