Statistics-And-Probability Question 634
Question: 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
Options:
A) 3 and 6
B) 2 and 6
C) 5 and 8
D) 1 and 7
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let the other two observations be a and b
$ \therefore $ mean $ =\frac{2+4+6+a+b}{5}\Rightarrow 4=\frac{12+a+b}{5} $
$ \Rightarrow a+b=8; $ Variance $ =\frac{1}{n}\sum{x^{2}}-{x^{-2}}=5.2 $
$ \Rightarrow \frac{1}{5}( 4+16+36+a^{2}+b^{2} )-16=5.2\Rightarrow a^{2}+b^{2} $ $ =50 $ From the options, it is clear that the two observations are 1 and 7.
BETA
