Question

The daily milk production of a herd of cows is assumed to be Normally distributed with a mean of 37 liters, and standard deviation of 5.6 liters. A) On what proportion of days is daily production less than 20.3 liters? Answer= (Round your answer to 3 decimal places.) B) On what proportion of days is production more than 49.1 liters? Answer= (Round your answer to 3 decimal places.)

Answer #1

Solution :

Given that ,

mean = = 37

standard deviation = = 5.6

A) P(x < 20.3) = P[(x - ) / < (20.3 - 37) / 5.6]

= P(z < -2.98) = 0.001

**Probability = 0.001**

B) P(x > 49.1) = 1 - P(x < 49.1)

= 1 - P[(x - ) / < (49.1 - 37) /5.6 )

= 1 - P(z < 2.16)

= 1 - 0.9846 = 0.015

Probability = 0.015

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