# Calculating probability of proposal favour

President Obama proposed the elimination of taxes on dividends paid to shareholders on the grounds that they result in double taxation. The earnings used to pay dividends are already taxed to the corporation. A survey by Dr Greg on this issue revealed that 47% of Americans favor the proposal. By political party, 64% of the democrats and 29% of the Republicans favor the proposal. Suppose a group of 250 Americans gather to hear a speech about the proposal.

What is the probability that at least half of the group is in favor of the proposal?

Attempt:

I tried solving it Using normal approximation to binomial but I'm not sure

The survey is a Bernouilli trial (yes/no, heads/tails, in favour of/against) so the binomial distribution applies.

The mean is np=250*0.47=117.5 and the standard deviation is s=sqrt(npq)=7.9.

I can't continue from here

• It is difficult to see how "probability" even applies here, given that the gathering of 250 Americans is guaranteed not to be random and its political makeup will depend heavily on the context, such as who is giving the speech.
– whuber
Oct 8, 2015 at 14:57
• If all of them are democrats then the $p_0$ should be set to 0.64 and if all of them are the republicans then $p_0$ should be set to 0.29. Oct 8, 2015 at 21:55

I think use normal approximation should be OK.

Let us treat $p_0=0.47$ as the parameter,

and $p=0.5$

Next we use CLT to calculate the critical value

$z=\frac{p-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}=\frac{0.5-0.47}{\sqrt{0.47*0.53/250}}=0.95$

(If you prefer use number of cases, just multiple $n$ for numerator and denominator at the same time)

$Probability=1-\Phi(0.95)=0.171$ for one sided $P$

So the probability that at least half of the group is in favor of the proposal is 17.1% when you use the normal approximation.

You also can use the exact distribution to calculate the Probability

$P(x\ge125)=P(x=125)+P(x=126)+....+P(x=250)=\binom{250}{125}p^{125}(1-p)^{250-125}+\binom{250}{126}p^{126}(1-p)^{250-126}+...+\binom{250}{250}p^{250}(1-p)^{250-250}$

Here $p=0.47$

You can use the following R code to calculate the probability

pr<-c()
for(i in 126:250){
pr[i-125]<-dbinom(i,250,0.47)

}
sum(pr)

#pr=0.1553555


Sorry, the above code miss $x=125$

dbinom(125,250,0.47)