Suppose a population $$A$$ has $$100$$ observations $$101,\,102,......,200$$ and another population $$B$$ has $$100$$ obsevrations $$151,\,152,......,250.$$ If $${V_A}$$ and $${V_B}$$ represent the variances of the two populations, respectively then $$\frac{{{V_A}}}{{{V_B}}}$$ is :
A.
$$1$$
B.
$$\frac{9}{4}$$
C.
$$\frac{4}{9}$$
D.
$$\frac{2}{3}$$
Answer :
$$1$$
Solution :
$$\sigma _x^2 = \frac{{\sum {d_i^2} }}{n}$$ (Here deviations are taken from the mean).
Since $$A$$ and $$B$$ both have $$100$$ consecutive integers, therefore both have same standard deviation and hence the variance.
$$\therefore \,\frac{{{V_A}}}{{{V_B}}} = 1$$ (As $${\sum {d_i^2} }$$ is same in both the cases)
Releted MCQ Question on Statistics and Probability >> Statistics
Releted Question 1
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