A) 2
B) 4
C) 3
D) 5
Correct Answer: B
Solution :
Here we are given \[{{n}_{1}}=100,{{\overline{x}}_{1}}=15\]and\[{{\sigma }_{1}}=3\] \[{{n}_{1}}+{{n}_{2}}=250,\overline{x}=15.6\]and\[\sigma =\sqrt{13.44}\] we want \[{{\sigma }_{2}}\]obviously, \[{{n}_{2}}=150,\] Now\[\overline{x}=\frac{{{n}_{1}}{{\overline{x}}_{1}}+{{n}_{2}}{{\overline{x}}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] \[15.6\times 250=100\times 15+150\times {{\overline{x}}_{2}}\] \[\Rightarrow \]\[{{\overline{x}}_{2}}=16\] Hence\[{{d}_{1}}-{{\overline{x}}_{1}}-\overline{x}=-0.6,{{d}_{2}}-{{\overline{x}}_{2}}-\overline{x}=0.4\] The variance of the combined group \[{{\sigma }^{2}}\] is given by the formula \[({{n}_{1}}+{{n}_{2}}){{\sigma }^{2}}={{n}_{1}}(\sigma _{1}^{2}+d_{1}^{2})+{{n}_{2}}(\sigma _{2}^{2}+d_{2}^{2})\] \[250\times 13.44=100(9+0.36)+150(\sigma _{2}^{2}+0.16)\] \[\Rightarrow \]\[{{\sigma }_{2}}=4\]
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