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(A) The population $A$ consists of $100$ consecutive integers: $101, 102, . . ., 200$.The population $B$ consists of $100$ consecutive integers: $151,

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(A) The population $A$ consists of $100$ consecutive integers: $101, 102, . . ., 200$.The population $B$ consists of $100$ consecutive integers: $151, 152, . . ., 250$.We know that the variance of a set of observations is independent of the change of origin. That is,if $y_i = x_i + c$,then $Var(y) = Var(x)$.Here,each observation in population $B$ can be written as $y_i = x_i + 50$,where $x_i$ are the observations of population $A$.Since the variance is invariant under the change of origin,$V_A = V_B$.Therefore,$\frac{V_A}{V_B} = 1$.

$x_i$	$60$	$61$	$62$	$63$	$64$	$65$	$66$	$67$	$68$
$f_i$	$2$	$1$	$12$	$29$	$25$	$12$	$10$	$4$	$5$

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:

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