# 52% of people want to ban smoking Use the normal approximation to estimate that over half of a sample size $n$ support the the ban

52% of people want to ban smoking.

Use the normal approximation to estimate that more than half of a given sample size $n$ support the the ban.

q=1-0.52=0.48

For $n=11, 101, 1001$

Are these steps correct:

1.) Finding the mean (p*n)

2.) multiplying by q (pnq), and taking the square root $\sqrt(pqn)$ to find std. dev.

3.)Now we want P(X$\ge$5.5) (5.5 is half or more of our sample size) and we need to subtract 0.5 I believe?

Assuming these steps are correct, what's next?

To find expressions like $P(X\geq x)$ where $X$ has normal distribution observe that $$P(X\geq x)=P(\sigma U+\mu\geq x)=P\left(U\geq\frac{x-\mu}{\sigma}\right)$$ where $U$ has standard normal distribution.

So actually for $z=\frac{x-\mu}{\sigma}$:$$P(X\geq x)=1-\Phi\left(z\right)$$

You can find this $z$ by substituting.

From here tables come in.

Also think of $\Phi(z_a)=a$ in your former question

• Thanks again! You've been a great help in tacking these problems! – user300011 May 25 '16 at 14:25
• You are very welcome. – drhab May 25 '16 at 14:25
• This might be overreaching, but do you mind helping me with another question I've posted? – user300011 May 25 '16 at 14:26
• I do not make any promises; too risky:-). Btw, where is it? – drhab May 25 '16 at 14:28
• – user300011 May 25 '16 at 14:29