The mean daily production of a herd of cows is assumed to be normally distribute

Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 38 liters, and standard deviation of 3.3 liters.

A) What is the probability that daily production is less than 43.1 liters?

Answer= (Round your answer to 4 decimal places.)

B) What is the probability that daily production is more than 32.7 liters?

Answer= (Round your answer to 4 decimal places.)

Warning: Do not use the Z Normal Tables...they may not be accurate enough since WAMAP may look for more accuracy than comes from the table.

Explanation / Answer

Ans:

mean = 38

standard deviation = 3.3

a)

z=(43.1-38)/3.3=1.5455

P(z<1.5455)=0.9389

b)

z=(32.7-38)/3.3=-1.606

P(z>-1.606)=P(z<=1.606)=0.9459

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