Question

2. (Textbook question 9.34) Calculate the mean and the standard deviation of the sampling distribution of possible sample proportions for each combination of sample size ( n) and population proportion (p). a. n = 64, p = 0.8 b. n = 256, p = 0.8

Answer 1

Answer:

a) $E(X) = n*p= 64*0.8= 51.2$

And the deviation is given by:

$\sigma = \sqrt{np(1-p)}=\sqrt{64*0.8*(1-0.8)}= 3.2$

b) $E(X) = n*p= 256*0.8= 204.8$

And the deviation is given by:

$\sigma = \sqrt{np(1-p)}=\sqrt{256*0.8*(1-0.8)}= 6.4$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=64, p=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by:

$E(X) = n*p= 64*0.8= 51.2$

And the deviation is given by:

$\sigma = \sqrt{np(1-p)}=\sqrt{64*0.8*(1-0.8)}= 3.2$

Part b

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=256, p=0.8)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The expected value is given by:

$E(X) = n*p= 256*0.8= 204.8$

And the deviation is given by:

$\sigma = \sqrt{np(1-p)}=\sqrt{256*0.8*(1-0.8)}= 6.4$