A) 5
B) 4
C) 6
D) 3.52
Correct Answer: B
Solution :
[b] Use \[{{\sigma }^{2}}=\frac{{{n}_{1}}(\sigma _{1}^{2}+d_{1}^{2})+{{n}_{2}}(\sigma _{2}^{2}+d_{2}^{2})}{{{n}_{1}}+{{n}_{2}}}\] where \[{{d}_{1}}={{m}_{1}}-a,{{d}_{2}}={{m}_{2}}-a,a\] being the mean of the whole group. Let \[{{m}_{2}}=\]mean of the second group \[\therefore 15.6=\frac{100\times 15+150\times {{m}_{2}}}{250}\Rightarrow {{m}_{2}}=16\] Thus, \[(100\times 9+150\times {{\sigma }^{2}})\] \[13.44=\frac{+100\times {{(0.6)}^{2}}+150\times {{(0.4)}^{2}}}{250}\] \[\Rightarrow \sigma =4\]
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