Let ${x_1}\;,\;{x_2}\;,\;.\;.\;.\;,{x_n}$ be $n$ observations, and let $\bar x$ be their arithmaetic mean and ${\sigma ^2}$ be the variance
Statement $-1$ :Variance of $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4{\sigma ^2}$ .
Statement $-2$: Arithmetic mean $2{x_1}\;,2\;{x_2}\;,\;.\;.\;.\;,2{x_n}$ is $4\bar x$.
Statement $-1$ is false, Statement $-2$ is true;
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not acorrect explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$
Statement $-1$ is true, Statement $-2$ is false
The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ is
For $(2n+1)$ observations ${x_1},\, - {x_1}$, ${x_2},\, - {x_2},\,.....{x_n},\, - {x_n}$ and $0$ where $x$’s are all distinct. Let $S.D.$ and $M.D.$ denote the standard deviation and median respectively. Then which of the following is always 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:
For the frequency distribution :
Variate $( x )$ | $x _{1}$ | $x _{1}$ | $x _{3} \ldots \ldots x _{15}$ |
Frequency $(f)$ | $f _{1}$ | $f _{1}$ | $f _{3} \ldots f _{15}$ |
where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and
$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be
The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2.$ then their new variance is equal to :