Suppose values taken by a variable $x$ are such that $a \le {x_i} \le b$, where ${x_i}$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ...n.$ Then..
$a \le {\rm{Var}}(x) \le b$
${a^2} \le {\rm{Var}}(x) \le {b^2}$
$\frac{{{a^2}}}{4} \le {\rm{Var}}(x)$
${(b - a)^2} \ge {\rm{Var}}(x)$
The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5,7,10,12,14,15,$ then the absolute difference of the remaining two observations is
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 $r$ be the range and ${S^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} $ be the $S.D.$ of a set of observations ${x_1},\,{x_2},\,.....{x_n}$, then
The diameters of circles (in mm) drawn in a design are given below:
Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]