Question

# The mean daily production of a herd of cows is assumed to be normally distributed with...

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 39 liters, and standard deviation of 7.6 liters.

A) What is the probability that daily production is less than 29.1 liters? Answer= (Round your answer to 4 decimal places.)

B) What is the probability that daily production is more than 28.7 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.

*that warning is my teacher's warning, not mine. I clearly don't know enough to have an opinion.

Mean = = 39

Standard deviation = = 7.6

A)

We have to find the probability that daily production is less than 29.1 liters.

That is we have to find P(X < 29.1)

For finding this probability we have to find z score.

$z =\frac{x-\mu}{\sigma} =\frac{29.1-39}{7.6}=-1.30$

That is we have to find P(Z < - 1.30)

P(Z < - 1.30) = 0.0968

( Using excel =NORMSDIST(-1.30))

B)

We have to find the probability that daily production is more than 28.7 liters.

That is we have to find P(X > 28.7)

For finding this probability we have to find z score.

That is we have to find P(Z > - 1.36)

P(Z > - 1.36) = 1 - P(Z < - 1.36) = 1 - 0.0869 = 0.9131

( Using excel =NORMSDIST(-1.36))

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