SOLUTION: Suppose the annual consumption of chicken mean is 20.84 pounds per person, and the standard deviation for the consumption of chicken per person is 9.193 pounds. The mean weight of

Question 1159739: Suppose the annual consumption of chicken mean is 20.84 pounds per person, and the standard deviation for the consumption of chicken per person is 9.193 pounds. The mean weight of chicken consumed for a sample of 200 randomly selected people is one value of many that form the sampling distribution of sample means.
a) What is the value for this sampling distribution?
b) What is the standard deviation of this sampling distribution?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
The mean of the sampling distribution is the same mean of 20.84
the sd is the sd of the population/sqrt(n)=9.193/sqrt(200)=0.6500 or 0.65
This means that if multiple samples of 200 people are taken, they will cluster around the mean of 20.84 pounds and have a sd of 0.65 pounds
RELATED QUESTIONS
Assume the mean annual consumption of peanuts is normally distributed, with a mean of 5.8 (answered by ewatrrr)

According to the data published by the U.S. Dept. of Agriculture, the average annual... (answered by rothauserc)

APPLES: the annual per capita consumption of fresh apples (in pounds) in the united... (answered by ikleyn)

Please help me solve this question, Assume the mean annual consumption of peanuts is... (answered by stanbon)

4. The USDA Economics and Statistics System at Cornell University maintains a Poultry... (answered by Glaviolette)

One last one hopefully maybe I just have the numbers wrong... American s ate an average (answered by stanbon)

Americans ate an average of 25.7 pounds of confectionary products each last year and... (answered by Boreal)

Americans ate an average of 25.7 pounds of confectionery products each last year and... (answered by Theo)

assume that in 1985 the annual consumption of pork was about 55 pounds per person an that (answered by stanbon,josgarithmetic)