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 36 liters, and standard deviation of 8.7 liters. a) What is the probability that daily production is less than 17.6 liters? First find the z-score. (Round your answer to 4 decimal places.) (Round your answer to 4 decimal places.) Answer- b) What is the probability that daily production is more than 29.5 liters? First find the z-score. z- (Round your answer to 4 decimal places.) Answer Round your answer to 4 decimal places)

Explanation / Answer

= 36 = 8.7

a) P(x < 17.6)

= P(z < (17.6 - 36) / 8.7)

= P(z < -2.1149)

= 0.0172.

z = -2.1149

Answer = 0.0172.

b) P(x > 29.5)

= P(z > (29.5 - 36) / 8.7)

= P(z > -0.7471)

= 0.7725.

z = -0.7471.

Answer = 0.7725.

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