A) 4
B) 5.66
C) 6.25
D) 8.33
Correct Answer: D
Solution :
\[N=20,A.M.=10\] \[A.M.=\frac{\sum{X}}{N};\] \[10=\frac{\sum{X}}{20};\sum{X}=200\] Now, the students with mean age of 15 years leave the class Sum of observation \[=\sum{{{X}_{1}}=A{{M}_{1}}\times {{N}_{1}}}\] \[A{{M}_{1}}=15,{{N}_{1}}=5,\sum{{{X}_{1}}}=75\] Remaining sum \[=\sum{X}-\sum{{{X}_{1}}}\] \[=200-75=125\] Remaining number of students \[=N-{{N}_{1}}\] \[=20-5=15\] Now, \[AM=\frac{125}{15}=8.33\]
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