PLS HELP 85 POINTS PLEASE PLEASE PLEASE!!! A rancher’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 How do the sample statistics compare to the population mean and standard deviation?

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.

saathvi19 saathvi19 · Answer 2 · 2020-07-07T23:33:26+02:00

Given:

250 sheep in a 40-acre pasture.

250/40 = 6.25 or 6 sheep per acre

n = 6

Sample 1: 4 → 4/6 = 0.67

Sample 2: 1 → 1/6 = 0.17

Sample 3: 9 → 9/6 = 1.50

Sample 1: 0.67(1-0.67) = 0.67(0.33) = 0.2211

Sample 2: 0.17(1-0.17) = 0.17(0.83) = 0.1411

Sample 3: 1.50(1-1.5) = 1.5(-0.5) = -0.75

Sample 1: 0.2211/6 = 0.03685

Sample 2: 0.1411/6 = 0.02352

Sample 3: -0.75/6 = -0.125

Sample 1: √0.03685 = 0.1919

Sample 2: √0.02352 = 0.1534

Sample 3: √-0.125 = invalid

z value 95% confidence 1.96.

Sample 1: 1.96 * 0.1919 = 0.3761 or 37.61% margin of error

Sample 2: 1.96 * 0.1534 = 0.3007 or 30.07% margin of error

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