Forty-seven percent of fish in a river are catfish. Imagine scooping out a simple random sample of 25 fish from the river and observing the sample proportion of catfish. What is the standard deviation of the sampling distribution? Determine whether the 10% condition is met.

Question options:

The standard deviation is 0.0998. The 10% condition is not met because there are less than 250 fish in the river.

The standard deviation is 0.0998. The 10% condition is met because it is very likely there are more than 250 fish in the river.

The standard deviation is 0.9002. The 10% condition is met because it is very likely there are more than 250 fish in the river.

The standard deviation is 0.9002. The 10% condition is not met because there are less than 250 fish in the river.

We are unable to 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 250 fish in the river

The standard deviation is 0.0998. The 10% condition is met because it is very likely there are more than 250 fish in the river

Explanation:

Formula to calculate standard deviation = √p(1-p)/n

Given p= population proportion

n= sample size

From the question,

We are given p(population proportion of catfish in the river)= 0.47

n(sample size of fish collected from river) = 25

The standard deviation using the formula above = √0.47(1-0.47)/25

= 0.0998

The 10% condition demands that a sample size be not less than 10% of the population

To check if the 10% condition was met, we know that a river will likely have up to 250 fish in it and so,

10% of 250= 0.10×250=25 fish

Therefore the option B chosen is correct

he can buy 4lbs. of live bait and 1lbs of natural bait

step-by-step explanation:

1 1/7 hope i

step-by-step explanation:

8/7 = (1 * 7 + 1) / 7 = 1 + 1/7 = 1 1/7