Assignment

Coefficients of variation of two distribution are $15$ and $20$ and their means are $20$ and $10$ respectively. If their standard deviations are ${\sigma }_1$ and ${\sigma }_2$ then

Given $\sum _{i=1}^{20}{a}_i=100,\sum _{i=1}^{20}{a}_i^2=600,\sum _{i=1}^{20}{b}_i=140,\sum _{i=1}^{20}{b}_i^2=1000$, where $a_i,b_i$ denotes length and weight of an observation. Then which is more varying?

The mean of $50$ observations were calculated as $4.04$ by Mohan who took by mistake $10$ instead of $8$ for one observation. The correct mean is

The standard deviation of first $10$ positive integral multiples of $3$ is equal to

The mean and standard deviations of $100$ observations were calculated as $40$ and $5.1$ respectively by a student who took by mistake $50$ instead of $40$ for one observation. The sum of correct mean and standard deviation is

If coefficient of variation of a distribution is $60\%$ and its standard deviation is $18$, then its arithmetic mean is

The variance of $x_1,x_2,x_3,\dots ,x_{100}$ is $13.75$ then variance of $5x_1,5x_2,5x_3,\dots ,5x_{100}$ is equal to

The mean and variance of seven observations are $8$ and $16$ respectively, if five of these are $2,4,10,12,14$ and other two are $a_1,a_2$ such that $a_1<a_2$ then $a_2^{a_2-a_1}$ is equal to