Answer:
The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405
Step-by-step explanation:
Central Limit Theorem:
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
18 percent are longer than 6.5 feet.
This means that [tex]p = 0.18[/tex]
A random sample of 90 sea turtles
This means that [tex]n = 90[/tex]
What is the standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90
[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.18*0.82}{90}} = 0.0405[/tex]
The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.0405
In this exercise we have to use the knowledge of proportionality to calculate the samples, so we have to:
The standard deviation of the sampling distribution of the sample proportion of sea turtles longer than 6.5 feet for samples of size 90 is of 0.00164
Using the knowledge about central limit theorem, we will have what the proportion will be;
For a balance between parts of whole p fashionable a sample of extent or bulk of some dimension n, the sampling allocation of the sample relative amount will exist nearly usual with mean and predictable difference, so we have that:
[tex]S=\sqrt{\frac{p(1-p)}{n} }\\p=0.18\\n=90\\S=\sqrt{\frac{0.18(1-0.18)}{90} }=0.00164[/tex]
See more about proportion at brainly.com/question/2548537
