Suppose a population $A$ has $100$ observations $101, 102, . . ., 200$ and another population $B$ has $100$ observations $151, 152, . . ., 250$. If $V_A$ and $V_B$ represent the variances of the two populations,respectively,then $V_A / V_B$ is:
- A$1$
- B$\frac{9}{4}$
- C$\frac{4}{9}$
- D$\frac{2}{3}$
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Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=b$,where $a > b$,be $5$ and $10$ respectively. Then the Variance of the observations $n+x_n$ for $n=1, 2, 3, 4, 5$ is:
DifficultJEE MAIN 2025
View SolutionThe variance of $50$ observations is $7$. Suppose that each observation in this data is multiplied by $6$ and then $5$ is subtracted from it. Then the variance of that new data is
The variance of $\alpha$,$\beta$,and $\gamma$ is $9$. Then,the variance of $5\alpha$,$5\beta$,and $5\gamma$ is:
Medium
View SolutionThe 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.] (in $\text{ mm}$)
| 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.] (in $\text{ mm}$)
Medium
View SolutionThe variance of the numbers $8, 21, 34, 47, \ldots, 320$ is . . . . . . .
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