QUESTION 1 1 points Save Answer Farmer Berton has a dairy farm with 25 cows. He

Question

QUESTION 1 1 points Save Answer Farmer Berton has a dairy farm with 25 cows. He measures all of his cows' milk production and calculates the average milk production per cow to be 72 lbs milk /cow each day. His local extension agent tells him that a study was recently conducted that found the distribution of all dairy cows' milk production (in the US) was normally distributed with a mean of 70 lbs milk / cow each day and a standard deviation of 10 Ibs milk/cow each day. What is the probability that a dairy farm has an average milk production less than that of Farmer Berton's farm? Express as a decimal and round to 3 decimal places. QUESTION 2 1 points Save Answer The average composite ACT score is 20.8 and the standard deviation is 4.8. A guidance counselor is interested in how her students' scores compare to the national average. A group of 144 students scored an average of 21. What is the probability of observing a mean greater than 21? Please express your answer as a decimal and round to 3 decimal places. QUESTION 3 1 pointsSave Answer The height of males in the US is normally distributed with a mean of 5'9" and a standard deviation of 3". The Fighting Illini basketball team has 17 team members, and the average height of the team is 6'5. Calculate the z-score you would use if you wanted to calculate any probabilities associated with this average height. Express your answer as a decimal and round your answer to 2 decimal places.

Explanation / Answer

a)

u=70, =10, x=72,n=25

z= (72-70)/(10/sqrt(10)) = 0.63

P(dairy farm has less avg production than Berton’s farm) = P(Berton’s farm has more avg production than dairy farm)

= P(Z>0.63) = 1 – 0.7357 = 0.264

b)

u=20.8, =4.8, x=21,n=144

z= (21-20.8)/(4.8/sqrt(144)) = 0.5

P(mean greater than 21) = P(Z>0.5) =1-0.6915 = 0.3085

c)

u=5’9” = 69”, =3”, x=6’5”=75”

z= (75-69)/3 = 2

P(mean greater than 6’5”) = P(Z>2) = 1-0.9772= 0.0228

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