Sampling Distribution of proportions

Suppose the population is infinite and propability of occurence of an event is p while probability of non-occuring is q. Consider all possible samples of size N drawn from this population , and for each sample determine the proportion P of success. Then a sampling distribution of proportions whose mean is \(\mu_p\) and standard deviation \(\sigma_p\) are given by folowing formulas \[\mu_p = p\] and \[\sigma_p=\sqrt{\frac{pq}{N}} = \sqrt{\frac{p(1-p)}{N}} \]

Lets take an example,the population may be all possible tosses of a fair coin in which the probability of the events “head” is p =1/2. P is proportion of head in N tosses.

Example Find the probability that 180 tosses of a fair coin (a)less than 40% or more than 60% will be head.

solution: We consider 180 tosses of the coin to be a sample from a infinite population of all possible tosses of coin.

probability of heads \(p=\frac{1}{2}\)
probability of not- heads \(q= 1- \frac{1}{2}\)

Less than 40% or more than 60% will be head i.e. total number heads less than 72 and more than 108

\(\mu_p = p = \frac{1}{2}\)

\(\sigma_p=\sqrt{\frac{pq}{N}}\)= 0.0372678

40% in standard units = \(\frac{0.4-0.5}{\sigma_p}\) = -2.6832816

60% in standard units = \(\frac{0.6-0.5}{\sigma_p}\) = 2.6832816

Area to left of -2.19 = pnorm(-2.19,0,1) =0.0142621
Area to righ of 2.19 = pnorm(2.19,0,1, lower.tail =FALSE) =0.0142621

So, less than 40% or more than 60% will be head = 0.02852

Important Although this result is accurate to significant figures,but we have not use the fact that the proporation is discrete variable. To account this,we subtract 1/2N from 0.4 and adding 1/2N to 0.6

curve(dnorm(x),xlim=c(-5,5),main='Standard Normal')
x1= (0.4 - 1/(2*180) - 0.50)/0.0456435
x2 = (0.6 + 1/(2*180) - 0.50)/0.0456435
codx <- c(-5,seq(-5,x1,0.1),x1)
cody <- c(0,dnorm(seq(-5,x1,0.1)),0)
polygon(codx,cody,col='red')
codx1 <- c(x2,seq(x2,5,0.1),5)
cody1 <- c(0,dnorm(seq(x2,5,0.1)),0)
polygon(codx1,cody1,col='red')

pnorm(x2,0,1, lower.tail =FALSE) + pnorm(x1,0,1)

## [1] 0.02433804

So, less than 40% or more than 60% will be head = 0.022