Answer: 0.0033
Step-by-step explanation:
Let x be a random variable that denotes the weights of cows.
Given: [tex]\mu = 483,\ \sigma=70[/tex]
maximum weight can be hold= 2840 pounds.
Mean weight = [tex]\frac{2840}{5}[/tex] = 568 pounds
The probability their total weight will be over the maximum allowed of 2840
= [tex]P(X>2840)[/tex]
[tex]P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{568-483}{\dfrac{70}{\sqrt{5}}})\\\\=P(z>2.715)\\\\=1-P(z<2.715)\\\\=1-0.9967=0.0033[/tex]
Hence, the required probability = 0.0033