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A vote on whether to allow the use of medical marijuana is being held. A polling company will survey $200$ individuals to measure support for the new law. If in fact $53\%$ of the population oppose the new law, use the normal approximation to the binomial, with a continuity correction, to approximate the probability that the poll will show a majority in favor?

Attempt

Let $X$ be the number of people who favor the law. The sample size is $n=200$. The probability of success $p = 0.47 $. Since $np = 94 >5$ with can approximate $X$ with a normal distribution. Notice, $\mu = np = 94$ is the number of people who favor the law. Thus, we want

\begin{align} P(X > 94) &\implies P(X \geq 94.5 ) \implies \\[2ex] &P\left( Z \geq \frac{94.5-\mu}{\sigma} \right) = P(Z \geq 0.0708) = 1 - \Phi (0.0708) = 0.4721\end{align}

But, this is far off from the answer which is $\boxed{0.1788}$. Is my approach correct? OR am I missing something?

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You want the probability that the poll will show a majority in favor, which happens if $X>100$. Otherwise, your solution appears to be correct.

Approximately we have that $X \sim \mathcal N (np,np(1-p))$ with $n=200$ and $p=0.47$, so with the continuity correction

$$ P(X>100.5) = P\left(Z>\frac{100.5-np}{\sqrt{np(1-p)}}\right)\approx P\left(Z>0.9209\right)\approx 0.1785,$$

where $Z \sim \mathcal N (0,1)$.

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