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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

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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Oct 8, 2015 at 14:57
  • $\begingroup$ 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. $\endgroup$
    – Deep North
    Oct 8, 2015 at 21:55

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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 $x=125$

Ok,just add the probability 

dbinom(125,250,0.47)
#0.03211816

So the exact probability will be about 18.5% no big difference with normal approximate one.

We did not use the information "By political party, 64% of the democrats and 29% of the Republicans favor the proposal". I am wondering what's that piece of information for?

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  • $\begingroup$ Thank you very much, Deep North. Appreciate it. Yeah the political party part does not matter. $\endgroup$
    – user274246
    Oct 8, 2015 at 14:59
  • $\begingroup$ How did you get p = 0.5 by the way please? $\endgroup$
    – user274246
    Oct 8, 2015 at 15:01
  • $\begingroup$ It was from "at least half of the group" $\endgroup$
    – Deep North
    Oct 8, 2015 at 21:50

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