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 32 liters, and standard deviation of 8.6 liters.

A) What is the probability that daily production is less than 57.1 liters? Answer= (Round your answer to 4 decimal places.)

B) What is the probability that daily production is more than 42.2 liters? 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.

Explanation / Answer

Mean = 32 liters

Standard deviation = 8.6 liters

A) P(X < 57.1) = P(Z < (57.1 - mean)/standard deviation)

= P(Z < (57.1 - 32)/8.6)

= P(Z < 2.9186)

= 0.9982

B) P(X > 42.2) = 1 - P(X < 42.2)

= 1 - P(Z < (42.2 - 32)/8.6)

= 1 - P(Z < 1. 1860)

= 1 - 0.8822

= 0.1178

P.S - Z value is not taken from z table, but online calculator is used to calculate z value rounded to four decimal places and the corresponding probability is also rounded to 4 decimal places)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.