Question

Georgia is a state with high production of peanuts on a yearly basis. Suppose 24 acres in the state of Georgia were randomly identified last year, and the peanut-yield rate, in pounds of peanuts harvested per acre, was recorded. The mean harvested weight of peanuts per acre for all counties in Georgia is known to be 3,200 pounds with a standard deviation of 465 pounds. The distribution of peanut-yield rate is approximately normal. (a) Calculate the mean (in pounds) of the sampling distribution of the sample mean. (b) Calculate the standard deviation (in pounds) of the sampling distribution of the sample mean. (Round your answer to three decimal places.) (C) What is the probability of obtaining a sample mean of less than 3070 pounds of peanuts harvested per acre based on a sample of 24 acres if the population mean truly is 3,200? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 3200

standard deviation = = 465

n = 24

a) = = 3200

b) = / n = 465 / 24 = 94.918

c) P( < 3070) = P(( - ) / < (3070 - 3200) / 94.918)

= P(z < -1.37)

Using z table

= 0.0853