I am dealing with a problem similar to what is described in Problem involving normal approximation to the binomial..
Assume that 30% of adults are smokers. Let $X$ be the number of adults in a random sample of size $n=100$. Use the normal approximation to the binomial to find: (i) the probability that between 10 and 20 (inclusive) of the sampled adults are smokers. (ii) the probability that the number of adults who smoke is at least 40% in another randomly selected sample of 100 adults.
I am confused about (ii) and that answer didn't help. I did the following:
$\mu = np = 100 * 0.40 = 40$
$\sigma^2 = npq = 100 * 0.40 * (1-0.40) = 100 * 0.40 * 0.60 = 24$
From that, I deduce that $\sigma=\sqrt(24)$
$P(X \geq 40) = 1 - P(X \lt 40) = 1 - P(Z \lt \frac{x + 0.5 - \mu}{\sigma})$
And I'm now officially lost! What's next?
Let $X \sim \mathcal{N}(np,\,np(1-p))$ be the normal approximation to the binomial.
So $X \sim \mathcal{N}(30,\,21)$.
$\frac{X - 30}{\sqrt{21}} = Z \sim \mathcal{N}(0,\,1)$
$$P(X \geq 40)=1-P(X\leq39)= 1-P\left(Z \leq\frac{39 + 0.5 - 30}{\sqrt{21}}\right)$$
Now you can just plug it into your calculator or use a lookup table.