Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 37 liters, and standard deviation of 4.7 liters. A) What is the probability that daily production is between 46.7 and 48.9 liters? Do not round until you get your your final answer.

Answer #1

Solution:

Given that,

= 37

= 4.7

A ) p (46.7 < x < 48.9 )

= p( 46.7 - 37 / 4.7 ) ( x - / ) < ( 48.9 - 37 / 4.7)

= p ( 9.7 / 4.7 < z < 11.9 /4.7 )

= p ( 2.06 < z < 2.53)

= p (z < 2.53 ) - p ( z < 2.06 )

Using z table

= 0.9943 -0.9803

= 0.0140

Probability = 0.0140

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