Answer
Sample of size 5:
$Mean=99.8$
$Median=100$
If 106 is accidentally recorded as 160:
$Mean=110.6$
$Median=100$
The mean increases substantially, but the median is not affected.
Sample of size 12:
$Mean=100.4$
$Median=101$
If 106 is accidentally recorded as 160:
$Mean=104.9$
$Median=101$
The mean increases, but the median is not affected.
Sample of size 30:
$Mean=100.6$
$Median=99$
If 106 is accidentally recorded as 160:
$Mean=102.4$
$Median=99$
The mean increases a little, but the median is not affected.
As the number of observations increases the mean becomes more resistant. The median is not affected by the number of observations.
Work Step by Step
Sample of size 5:
In ascending order: 92, 98, 100, 103, 106
$Mean=\frac{92+98+100+103+106}{5}=99.8$
In a sample of size 5 the median is the third observation in ascending order.
$Median=100$
If 106 is accidentally recorded as 160:
In ascending order: 92, 98, 100, 103, 160
$Mean=\frac{92+98+100+103+160}{5}=110.6$
$Median=100$
Sample of size 12:
In ascending order: 70, 83, 92, 98, 98, 100, 102, 103, 106, 108, 121, 124
$Mean=\frac{70+83+92+98+98+100+102+103+106+108+121+124}{12}=100.4$
In a sample of size 12 the median is the mean between the sixth and the seventh observations in ascending order.
$Median=\frac{100+102}{2}=101$
If 106 is accidentally recorded as 160:
In ascending order: 70, 83, 92, 98, 98, 100, 102, 103, 108, 121, 124, 160
$Mean=\frac{70+83+92+98+98+100+102+103+108+121+124+160}{12}=104.9$
$Median=\frac{100+102}{2}=101$
Sample of size 30:
In ascending order: 70, 72, 81, 83, 87, 88, 89, 90, 92, 93, 97, 97, 98, 98, 98, 100, 102, 102, 103, 103, 103, 106, 107, 108, 114, 121, 121, 124, 130, 140
$Mean=\frac{70+72+81+83+87+88+89+90+92+93+97+97+98+98+98+100+102+102+103+103+103+106+107+108+114+121+121+124+130+140}{30}=100.6$
In a sample of size 30 the median is the mean between the 15th and the 16th observations in ascending order.
$Median=\frac{98+100}{2}=99$
If 106 is accidentally recorded as 160:
In ascending order: 70, 72, 81, 83, 87, 88, 89, 90, 92, 93, 97, 97, 98, 98, 98, 100, 102, 102, 103, 103, 103, 107, 108, 114, 121, 121, 124, 130, 140, 160
$Mean=\frac{70+72+81+83+87+88+89+90+92+93+97+97+98+98+98+100+102+102+103+103+103+107+108+114+121+121+124+130+140+160}{30}=102.4$
$Median=\frac{98+100}{2}=99$
