First, we find the class marks of the given data as follows.
weight(in~ g) & Number of packets(fi)Class marks (xi)deviation(di=xi−a)fidi200−20113200.5−3−39201−20227201.5−2−54202−20318202.5−1−18203−20410a=203.500204−2051204.511205−2061205.522N=∑fi=70∑fidi=−108
Here, (assumed mean) a=203.5
and (class width) h=1
By assumed mean method,
Mean (¯x)=a+∑fidi∑fi=203.5−10870=203.5−1.54=201.96
Hence, the required mean weight is 201.96 g.