1

So this is a question about Normal Distribution (correct me if I am wrong), but I don't know how to solve it as I don't have variance and I'm not sure about mean. Help is highly appreciated.

Assume we have 2 candidates A and B applying for some position, and we estimate that 53% of the voters prefer the candidate B. We organize a poll, asking randomly chosen 100 voters about their preferences. What is the (approximate) probability that at least the half of that polled persons will favor the candidate A? Note: You can take Φ(0.601) ≈ 0.726.

Tigran Minasyan
  • 59
  • The mean of a binomial distribution with parameters $n$ and $p$ is $np$ while the variance is $np(1-p)$ – Henry May 22 '19 at 18:53
  • @Henry I know that... isn't this normal distribution? – Tigran Minasyan May 22 '19 at 18:54
  • 1
    It is explicitly a binomial distribution, however normal distributions act very similarly and so may be used to approximate the results for the binomial distribution, resulting in much easier calculations while sacrificing a little bit of accuracy. (With numbers like yours, expect the true answer to be off from the approximated answer by something on the order of magnitude of $0.0001$ or so) – JMoravitz May 22 '19 at 18:57
  • @JMoravitz thanks, but what about the solution? That is ultimately what I need, I know that binomial and normal are similar – Tigran Minasyan May 22 '19 at 18:58
  • 2
    In order to get started using a normal distribution to approximate your binomial distribution, the first steps are to decide which normal distribution to use. You want it to have the same mean and the same variance as your original binomial distribution, hence @Henry's comment about how to find the mean and variance of your actual binomial distribution. – JMoravitz May 22 '19 at 18:58
  • 1
    To continue, you can refer to similar problems such as this one or this one, refer to the wikipedia page here, self-help sites like here, or youtube lectures like here. – JMoravitz May 22 '19 at 19:04
  • Also, if more then half voters prefer B, probability of at least half voters favor A should be less than $0.5$. Is there a typo somewhere? – mihaild May 22 '19 at 19:06
  • @mihaild no, no typo. We are only polling a small number of the total population, not the whole population – JMoravitz May 22 '19 at 19:09
  • @JMoravitz but median of binomial distribution is $np$ if it's integer, and it is in this case. So probability of B getting less than expected number of votes (ie less than 53) should be less than $0.5$. – mihaild May 22 '19 at 19:12
  • @JMoravitz I just tried to solve it with $\mu=0.53100, \sigma=0.53100*0.47$ and $P(X\leq50)=P(Z\leq\frac{50-\mu}{\sigma^2})$, and the fraction was not equal to $0.601$... I think something is wrong here. – Tigran Minasyan May 22 '19 at 19:16
  • @TigranMinasyan $np(1 - p)$ is variance ($\sigma^2$), not standard deviation ($\sigma$). – mihaild May 22 '19 at 19:20
  • @mihaild I didn't write $\sigma^2$, that was a typo, I wrote just $\sigma$ – Tigran Minasyan May 22 '19 at 19:23

1 Answers1

1

Every vote can be seen as a bernoulli random variable, with $p=1-0.53=0.47$. Therefore the sum of the 100 votes for candidate $A$ is binominal distributed as $S\sim Bin(100, 0.47)$. with $\mu=47$ and $\sigma=\sqrt{100\cdot 0.47\cdot 0.53}$

Now it is asked for $P(S \geq 50)$. We can use the converse probability here. $P(S \geq 50)=1-P(S\leq 49)$

By applying $\text{central limit}$ theorem we can approximate the binomial distribution by the normal distribution.

$P(S \geq 50)=1-P(S\leq 49)\approx 1-\Phi \left( \frac{49-47}{\sqrt{100\cdot 0.47\cdot 0.53}} \right)=1-\Phi(0.401)$

Now we have the problem that the lecturer expect a different value, since it was calculated $P(S \geq 50)=1-P(S\leq 50)\approx 1-\Phi \left( \frac{50-47}{\sqrt{100\cdot 0.47\cdot 0.53}} \right)=1-\Phi(0.601)$

But this is wrong since $P(X\geq x)=1-P(X\leq x-1)$, if $X$ is binomial distributed. It´s up to you how you deal with this situation.

callculus42
  • 30,550