2
$\begingroup$

I have a statistics exercise, with a stochastic variable $X$ that has a $\text{Binomial}(64,p)$ distribution. P is unknown, and the goal is to find it.

In one of the sub-exercises, I have to approach the following:

$$P\left(\left|\frac{x}{64} - p\right| \le 0.1\right)$$

However, I simply have no idea where to even start. Can anyone attempt to explain how to approach a problem like this?

$\endgroup$

1 Answer 1

2
$\begingroup$

Lets approximate the $\dfrac{\text{Binomial}(64,p)}{64}$ with the normal distribution $N\bigl(p,\dfrac{p(1-p)}{64}\bigl)$, as the $\text{Binomial}(64,p)$ is just a repetition of identical expetiments. Then $\sqrt{\dfrac{64}{p(1-p)}}\bigl(\dfrac{\text{Binomial}(64,p)}{64}-p\bigr)$ is approximated as $N(0,1)$. And we come to

$P\bigl(\bigl|\frac{X}{64}-p\bigr|<0.1\bigr) = P\bigl( \sqrt{\frac{64}{p(1-p)}}\bigl|\frac{X}{64}-p\bigr|<\sqrt{\frac{64}{100p(1-p)}}\bigr) = Ф\bigl(\sqrt{\frac{64}{100p(1-p)}}\bigl)-Ф\bigl(-\sqrt{\frac{64}{100p(1-p)}}\bigl).$

$\endgroup$
2
  • $\begingroup$ That makes sense. I just don't follow the last step you're making; Can you explain why $P(|\frac X {64} - p| < 0.1)$ is equal to the phi of 0.1 minus the phi of -0.1? $\endgroup$
    – Erik S
    Jun 24, 2015 at 14:06
  • $\begingroup$ @ErikDolor, sorry, mistakes are fixed $\endgroup$ Jun 24, 2015 at 15:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .