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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