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The mean daily production of a herd of cows is assumed to be normally distributed with...

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 30 liters, and standard deviation of 10 liters. A) What is the probability that daily production is between 20.8 and 52 liters? Do not round until you get your your final answer. Answer= (Round your answer to 4 decimal places.)

Math   Statistics-And-Probability   
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Answer #1

µ = 30

sd = 10

                                      

                                       = P(-0.92 < Z < 2.2)

                                       = P(Z < 2.2) - P(Z < -0.92)

                                       = 0.9861 - 0.1788

                                       = 0.8073

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