Question

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.

Answer 1

Answer:

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.