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 33 liters, and standard deviation of 9.9 liters.

A) What is the probability that daily production is less than 9 liters?

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

B) What is the probability that daily production is more than 39.6 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

Without using the Z-table, we can solve this question in MS Excel.

A. The probability that daily production is less than 9 liters can be found using the EXcel fornula =NORM.DIST(9,33,9.9,TRUE)

Using the above, P(X<9) = 0.0076

B. The probability that daily production is more than 39.6 liters, P(X>39.6) = 1 - P(X<39.6) = 1 - (=NORM.DIST(39.6,33,9.9,TRUE)) = 1 - 0.7475 = 0.2525

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