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:
$18\left(1+\frac{1}{\sqrt{3}}\right)$
$34$
$2\left(9+\frac{8}{\sqrt{7}}\right)$
$25$
The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
Suppose a population $A $ has $100$ observations $ 101,102, . . .,200 $ and another population $B $ has $100$ observation $151,152, . . .,250$ .If $V_A$ and $V_B$ represent the variances of the two populations , respectively then $V_A / V_B$ 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?
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