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i'm stuck on this problem, so can somebody show me how to do it? !
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
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 = 6
sample proportion: signified by ρ
Sample 1: 4 → 4/6 = 0.67
Sample 2: 1 → 1/6 = 0.17
Sample 3: 9 → 9/6 = 1.50

multiply the sample proportion by 1-ρ
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

divide the result by n. n = 6
Sample 1: 0.2211/6 = 0.03685
Sample 2: 0.1411/6 = 0.02352
Sample 3: -0.75/6 = -0.125

square root of the quotient to get the standard error.
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

The correct answer is $90