A) 6 : 7
B) 10 : 3
C) 4 : 9
D) 5 : 8
Correct Answer: C
Solution :
\[\mu =5;\,\,\,variance\,\,=\,\,9.20\] \[\mu =\frac{1+3+8+{{x}_{1}}+{{x}_{2}}}{4}=\,5\Rightarrow {{x}_{1}}+{{x}_{2}}=13\] \[9.20\,=\,\frac{1}{n}\Sigma {{x}_{i}}^{2}-{{\mu }^{2}}\,\,\Rightarrow \,\,\Sigma {{x}_{i}}^{2}\,=\,(9.20+25)5\] \[1+9+64+{{x}_{1}}^{2}+{{x}^{2}}_{2}=171\] \[{{x}_{1}}^{2}+{{x}^{2}}_{2}=97\,\,\Rightarrow \,\,{{x}_{1}}^{2}+{{(13-{{x}_{1}})}^{2}}=97\] \[\left( {{x}_{1}}-9 \right)\left( {{x}_{1}}-4 \right)=0\] \[{{x}_{1}}=9;\,\text{ }{{x}_{2}}=4\] \[{{x}_{1}}=4;\,\text{ }{{x}_{2}}=9\] \[{{x}_{1}}:{{x}_{2}}=4:9\,\,or\text{ }9:4\]
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