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16 January, 20:36

Fifty-eight percent of the fish in a large pond are minnows. Imagine scooping out a simple random sample of 20 fish from the pond and observing the sample proportion of minnows. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met. The standard deviation is 0.1104. The 10% condition is not met because there are less than 200 minnows in the pond. The standard deviation is 0.1104. The 10% condition is met because it is very likely there are more than 200 minnows in the pond. The standard deviation is 0.8896. The 10% condition is met because it is very likely there are more than 200 minnows in the pond. The standard deviation is 0.8896. The 10% condition is not met because there are less than 200 minnows in the pond. We cannot determine the standard deviation because we do not know the sample mean. The 10% condition is met because it is very likely there are more than 200 minnows in the pond.

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  1. 16 January, 23:53
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    The standard deviation is 0.1104. The 10% condition is met because it is very likely there are more than 200 minnows in the pond.

    Step-by-step explanation:

    The standard deviation for a proportion is:

    σ = √ (pq/n)

    where p is the proportion, q is 1-p, and n is the sample size.

    σ = √ (0.58 * 0.42 / 20)

    σ = 0.1104

    Since the pond is large, there's probably more than 200 minnows, so the 10% condition is met.
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