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
$1$
$\frac{9}{4}$
$\frac{4}{9}$
$\frac{2}{3}$
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.]
The average marks of $10$ students in a class was $60$ with a standard deviation $4$, while the average marks of other ten students was $40$ with a standard deviation $6$. If all the $20$ students are taken together, their standard deviation will be
The mean and variance of seven observations are $8$ and $16$, respectively. If $5$ of the observations are $2, 4, 10, 12, 14,$ then the product of the remaining two observations is
The data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |