Question: How do I calculate the variance for a binomial distribution? The question in the Learning journal is: 3) Describe (do not directly quote any source)

How do I calculate the variance for a binomial distribution?

The question in the Learning journal is:

3) Describe (do not directly quote any source) the difference between the distribution of a sample and the sampling distribution.Use an example in which the original population has a binomial distribution.You will probably use concepts from the book or another source, so be sure to cite any concepts that come from such sources (even if you paraphrase).

Answer:

We have a population of x values.When we take a sample (n) from this population, calculate a simple mean from that sample, and repeat the process an infinite number of times, the distribution of this sample means is known as the sampling distribution of the mean.In other words the distribution of a sample mean is the process prior to the gathering of a large number of sample means which leads to the sampling distribution (Yakir, 2011).

A sampling distribution can only be described by a binomial distribution where the population is much larger than the sample size.So we can take as an example of a binomial distribution for a large population who suffers from high blood pressure in the eye leading to glaucoma with a 15% probability of becoming blind (Glaucoma, 2019).The conditions of a binomial distribution must be :

a.The number of observations n is fixed.

b.Each observation is independent.

c.Each observation represents one of two outcomes ("success" or "failure").

d.The probability of "success" p is the same for each outcome.

Once these conditions are fulfilled, the binomial distribution can be calculated in R with the formula B(n,p).

A sample study of 100 individuals is carried out where the random variable describing the number of individuals who will contract the disease is distributed as

B(n=100, p=0.15;

B(100,0.15).

In R

> n=100

>p=0.15

[1] 15

Then we compute the variance as follows:

And this is where I got stuck

jQuery22407316394083304296_1603055065817??

Var(n*p(1-p))but it doesn't work

Finally, we can compute the standard deviation

>square root(n*p*(1-p))

[1] 3.570714

But I think the standard deviation is also wrong. Please can you help? Thank you.

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