I have my statistics exam quite soon and i came upon this question :
At the last referendum, $40\%$ of the Italian population supported the
constitutional reform. If a random sample of size $n = 200$ is drawn, which
is the probability of observing at least $100$ people who voted YES?
Searching through my online notes, i found out that the explanation that the professor gave is this:
If $X$ is the random variable which is equal to one if the unit voted YES, then it has a Bernoulli distribution with success probability equal $0.40$. Due to the large sample size, the Normal approximation holds.
Therefore, $\hat p \sim \mathcal N(0.40, 0.012)$ and
$Pr(\hat p > 0.50) = Pr(Z > 0.1/0.0346) = Pr(Z > 2.89) = 1 − Pr(Z < 2.89) = 0.0019$.
I do not understand how he went through this ... can somebody help me? Starting from the first part of the explanation , where he states:
$\hat p\sim \mathcal N(0.40, 0.012)$.
Where is the $0.012$ coming from ?
Thanks!
The key point is that the random variable is the proportion. First we have the sum of 200 i.i.d. Bernoulli random variables $S_{200}=X_1+X_2+\ldots+X_{200}$, where $X_i\sim Ber(0.4)$. Therefore $S_{200}$ is binomial distributed as $Bin(200, 0.4)$. The variance is just $n\cdot p\cdot (1-p)=200\cdot 0.4\cdot 0.6$. So far so good. The standardized random variable Z (mean=$0$, variance=$1$) is $$Z=\frac{S_{200}-80}{\sqrt{48}} \ \dot \sim \ \mathcal N(0,1)$$
But the twist is that they now calculate the random variable as proportion. For this purpose they divide the numerator and the denominator by $n=200$.
Thus $Z=\frac{\hat p-p}{\sqrt{p\cdot (1-p)/n}}=\frac{S_{200}/200-80/200}{\sqrt{48/200^2}}\ \dot \sim \ \mathcal N(0,1)$
with $80/200=0.4$ and $\sqrt{48/200^2}=\sqrt{0.0012}$
Note that I have one more $0$ under the square root.
And we are looking for the probability where $\hat p=S_{200}/200\geq 0.5$. Thus we have
$$P\left(S_{200}/200 \geq 0.5 \right)=1-P\left(S_{200}/200 \leq 0.5 \right)$$
$$\approx 1-\Phi\left( \frac{0.5-0.4}{\sqrt{0.0012}} \right)=1-\Phi\left( \frac{0.1}{0.0346} \right)$$
Edit:
In the comment of BruceET the fiqures makes me thinking. One important problem in the given solution is that it is not regarded that the binomial distribution is discrete. That means that $P(X\geq 100)=1-P(X\leq 99)$. So we don´t use the $100$ if we apply the converse probability.
Without the continuity correction factor we have
$$\approx 1-\Phi\left( \frac{0.495-0.4}{\sqrt{0.0012}} \right)=1-\Phi\left( \frac{0.095}{0.0346} \right)=1-\Phi\left(2.745665 \right)=0.003020$$
With the continuity correction factor we have
$$\approx 1-\Phi\left( \frac{0.4975-0.4}{\sqrt{0.0012}} \right)=1-\Phi\left( \frac{0.0975}{0.0346} \right)=1-\Phi\left(2.8179191 \right)=0.002417$$
If we use the binomial distribution we get $0.00263540336$