Answer:
The sample of farms required should be 550 large.
Step-by-step explanation:
Let the sample =K,
mean=np=100
Standard deviation=[tex]\sqrt{npq}[/tex]=500
probability =Z=0.90
from the normal probability formula,
Z=[tex]\frac{K-np}{\sqrt{npq}}[/tex]
0.9=[tex]\frac{K-100}{500}[/tex]
500 X 0.9=K-100
450=K-100
K=450+100=550
Hence, the sample of farms required should be 550 large.