The mean of five observations is $$4$$ and their variance is $$5 \cdot 2.$$ If three of these observations are $$2,\,4$$ and $$6$$, then the other two observations are :
A.
$$3$$ and $$5$$
B.
$$2$$ and $$6$$
C.
$$5$$ and $$8$$
D.
$$1$$ and $$7$$
Answer :
$$1$$ and $$7$$
Solution :
Let the other two observations be $$'a$$ and $$'b'$$
$$\eqalign{
& \therefore {\text{ Mean}} = \frac{{2 + 4 + 6 + a + b}}{5} \cr
& \Rightarrow 4 = \frac{{12 + a + b}}{5} \cr
& \Rightarrow a + b = 8\,; \cr
& {\text{Variance}} = \frac{1}{n}\sum {{x^2} - {x^{ - 2}}} = 5.2 \cr
& \Rightarrow \frac{1}{5}\left( {4 + 16 + 36 + {a^2} + {b^2}} \right) - 16 = 5.2 \cr
& \Rightarrow {a^2} + {b^2} = 50 \cr} $$
From the options, it is clear that the two observations are $$1$$ and $$7.$$
Releted MCQ Question on Statistics and Probability >> Statistics
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