1

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?

JoeyNYC
  • 11
  • $p=0.3$ and not $0.4$ in your calculations for $\mu$ and $\sigma^2$? – Shuri2060 Jul 02 '17 at 00:01
  • So I should have $\mu=30$ and $\sigma^2=21$? OK... but what's next? – JoeyNYC Jul 02 '17 at 00:04
  • $x=40$ in your last step. that gives you the $z$. – imranfat Jul 02 '17 at 00:17
  • The given problem is Binomial so $P(X\geq40)=1-P(X\leq39)$ and so in your last step the little $x$ becomes actually $39$. Think about it in terms of "bars". You need to include in your probability the 40th bar and above, so you you need to subtract the 39th bar and below. The 39th and 40th bar "meet" at $39.5$ – imranfat Jul 05 '17 at 20:28

1 Answers1

1

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.

imranfat
  • 10,029
Shuri2060
  • 4,353