Answer:
The sample of farms required should be 550 large.
Explanation:
Let the sample =K,
mean=np=100
Standard deviation=
=500
probability =Z=0.90
from the normal probability formula,
Z=

0.9=

500 X 0.9=K-100
450=K-100
K=450+100=550
Hence, the sample of farms required should be 550 large.