Question
Asked Dec 9, 2019
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The mean daily production of a herd of cows is assumed to be normally distributed with a mean of
35 liters, and standard deviation of 6.7 liters.
A) What is the probability that daily production is between 25.6 and 41.6 liters? Do not round until
you get your your final answer.
(Round your answer to 4 decimal places.)
Answer=
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.
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The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 35 liters, and standard deviation of 6.7 liters. A) What is the probability that daily production is between 25.6 and 41.6 liters? Do not round until you get your your final answer. (Round your answer to 4 decimal places.) Answer= 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.

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Step 1

Given:

X:Daily production of herd of cows.

Population mean = μ = 35

Population standard deviation = σ = 6.7

Step 2

A)We have to find P(25.6 < X < 41.6) = …?

The corresponding z-values needed to be computed are

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Z, 25.6 – 35 Z, 6.7 Z =-1.403 And X, -H Z, 41.6– 35 Z, 6.7 Z, = 0.9851

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Step 3

Therefore, w...

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P(25.6 <X< 41.6) = P(-1.403 <Z<0.9851) P(25.6 <X< 41.6) = P(Z< 0.9851) – P(Z<-1.403) P(25.6 <X<41.6) =0.8377 – 0.0803 ...Using excel =NORMSDIST(0.9851) and =NORMSDIST(-1.403) P(25.6 <X< 41.6) = 0.7574

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