A) 2 and 9
B) 3 and 8
C) 4 and 7
D) 5 and 6
Correct Answer: C
Solution :
Let the two unknown items be x and y, then Mean \[=4\Rightarrow \frac{1+2+6+x+y}{5}=4\] Þ \[x+y=11\] .....(i) and variance = 5. 2 Þ \[\frac{{{1}^{2}}+{{2}^{2}}+{{6}^{2}}+{{x}^{2}}+{{y}^{2}}}{5}-{{(\text{mean})}^{2}}=5.2\] \[41+{{x}^{2}}+{{y}^{2}}=5[5.2+{{(4)}^{2}}]\] \[41+{{x}^{2}}+{{y}^{2}}=106\] \[{{x}^{2}}+{{y}^{2}}=65\] .....(ii) Solving (i) and (ii) for \[x\] and y, we get \[x=4,y=7\] or \[x=7,y=4\].
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