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.

The standard deviation should be 0.1104. The 10% condition should be met since it is very likely there are more than 200 minnows in the pond.

Calculation of the standard deviation:

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.

So, The standard deviation for a proportion is:

[tex]= \sqrt (pq\div n)[/tex]

Here

p is the proportion,

q is 1−p,

and n is the sample size.

So,

[tex]= \sqrt (0.58 \times 0.42 \div 20)[/tex]

= 0.1104

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

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