Arancher’s herd of 250 sheep grazes over a 40-acre pasture. he would like to find out how many sheep are grazing on each acre of the pasture at any given time, so he has some images of the pasture taken by the state department of agriculture’s aerial photography division. here are three samples of the images. sample 1. | 4 sample 2. | 1 sample 3. | 9 1.] what can the rancher conclude from these samples about how many sheep graze on each acre of the 40-acre pasture? 2.] if the sheep were equally "spread out" across all of the 40 acres, how many sheep would you expect to find on average on each acre? 3.] how do the sample statistics compare to the population mean and standard deviation? 4.] what margin of error is reasonable for these samples?

Given:250 sheep in a 40-acre pasture.Number of sheep grazing in each acre.
250/40 = 6.25 or 6 sheep per acre
n = 6sample proportion: signified by ρSample 1: 4 → 4/6 = 0.67Sample 2: 1 → 1/6 = 0.17Sample 3: 9 → 9/6 = 1.50
multiply the sample proportion by 1-ρSample 1: 0.67(1-0.67) = 0.67(0.33) = 0.2211Sample 2: 0.17(1-0.17) = 0.17(0.83) = 0.1411Sample 3: 1.50(1-1.5) = 1.5(-0.5) = -0.75
divide the result by n. n = 6Sample 1: 0.2211/6 = 0.03685Sample 2: 0.1411/6 = 0.02352Sample 3: -0.75/6 = -0.125
square root of the quotient to get the standard error.Sample 1: √0.03685 = 0.1919Sample 2: √0.02352 = 0.1534Sample 3: √-0.125 = invalid
z value 95% confidence 1.96.
Sample 1: 1.96 * 0.1919 = 0.3761 or 37.61% margin of errorSample 2: 1.96 * 0.1534 = 0.3007 or 30.07% margin of error

None liner bro because