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
$1$
$\frac{9}{4}$
$\frac{4}{9}$
$\frac{2}{3}$
If both the means and the standard deviation of $50$ observations $x_1, x_2, ………, x_{50}$ are equal to $16$ , then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, …., (x_{50} - 4)^2$ is
Let $y_1$ , $y_2$ , $y_3$ ,..... $y_n$ be $n$ observations. Let ${w_i} = l{y_i} + k\,\,\forall \,\,i = 1,2,3.....,n,$ where $l$ , $k$ are constants. If the mean of $y_i's$ is is $48$ and their standard deviation is $12$ , then mean of $w_i's$ is $55$ and standard deviation of $w_i's$ is $15$ , then values of $l$ and $k$ should be
Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
The mean and standard deviation 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. What are the correct mean and standard deviation?