The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 36 liters, and standard deviation of 11 liters.

Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 36 liters, and standard deviation of 11 liters.

A) What is the probability that daily production is between 53.7 and 64.7 liters? Do not round until you get your your final answer.

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.

Answer 1

a) x daily production in liters, mean 36, SD 11.

P(53.7 < x < 64.7)

transform to standard normal (z)

P( (53.7 - 36)/11 < z < (64.7 - 36)/11) = P(1.61 < z < 2.61) = P(z < 2.61) - P(z < 1.61) = 0.9955 - 0.9463 = 0.0492