The correct option is A 20
Let x1,x2,x3,x4,x5 be the height of the five students, so
5∑i=1xi=5×150=750
and 5∑i=1x2i5−1502=18
⇒5∑i=1x2i=5×(18+1502)
Let x6 be the height of the new student.
So x6=156 cm
Now
6∑i=1xi=750+156=906 and 6∑i=1x2i=5×(18+1502)+1562
Then, new variance=6∑i=1x2i6−⎛⎜
⎜
⎜
⎜⎝6∑i=1xi6⎞⎟
⎟
⎟
⎟⎠2=5×(18+1502)+15626−(9066)2=5×(18+1502)+15626−(151)2=20
Alternate solution :
From given data ,
Let ¯¯¯¯¯x1=150, ¯¯¯¯¯x2=156
σ21=18, σ22=0 (considering only one observation
n1=5,n2=1
¯¯¯x=5(150)+1566=151
Now using combined variance formulae,
σ2=1n1+n2[n1(σ21+d21)+n2(σ22+d22)]
where d1=¯¯¯x−¯¯¯¯¯x1=151−150=1
d2=¯¯¯x−¯¯¯¯¯x2=151−156=−5σ2=16[5(18+1)+1(02+25)]⇒σ2=1206=20