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 10.1 liters. A) What is the probability that daily production is less than 60.6 liters? Answer= (Round your answer to 4 decimal places.) B) What is the probability that daily production is more than 22.2 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.

Answer 1

A) The probability that daily production is less than 60.6 liters is 0.9926

B) The probability that daily production is more than 22.2 liters is 0.9141

The complete solution of above question are as below

The values provided in the above question are

Mean = = 36

standard deviation = = 10.1

A) We find the probability that daily production is less than 60.6 liters

Consider, X = daily production

----------(1)

We convert above X into Z using following formula

------------(2)

Using equation (2) in equation (1) we get

We find above probability using Following Excel function

=NORMSDIST(z)

Here, z = 2.435643564

=NORMSDIST( 2.435643564) then press Enter

=0.992567338 0.9926 (Round answer up to 4 decimal places.)

The probability that daily production is less than 60.6 liters is 0.9926

B) We find the probability that daily production is more than 22.2 liters

Consider, X = daily production

----------(1)

We convert above X into Z using following formula

------------(2)

Using equation (2) in equation (1) we get

We find above probability using Following Excel function

=1 - NORMSDIST(z)

=1 - NORMSDIST(-1.366336634) then press Enter

= 0.914083336 0.9141 (Round answer up to 4 decimal places.)

The probability that daily production is more than 22.2 liters is 0.9141

Summary :-

A) The probability that daily production is less than 60.6 liters is 0.9926

B) The probability that daily production is more than 22.2 liters is 0.9141