1
$\begingroup$

a fair die is rolled 100 times. What is the probability that "6" appears more than 15 times? Use the normal approximation with continuity correction.

I've found the mean to be $100/6$ or $50/3$ and $\sigma$ as $3.727$ but I'm unsure of how to use continuity correction? Do I do a $Z$-Test ?

$\endgroup$
1
$\begingroup$

Let random variable $X$ be the number of $6$'s. Then $X$ has mean $\mu=100\cdot \frac{1}{6}$ and variance $\sigma^2=100\cdot \frac{1}{6}\cdot\frac{5}{6}$.

The random variable $X$ is a sum of a fairly large number of reasonably nice independent identically distributed random variables, so the cdf of $X$ is reasonably well approximated by the cdf of a normal $W$ with the same mean and variance.

The probability that $X\le 15$ is approximately the probability that $W\le 15$. One can expect a better approximation by the probability that $W\le 15.5$. (This is the continuity correction.) We have $$\Pr(W\le 15.5)=\Pr\left(Z\le \frac{15.5-\mu}{\sigma}\right),$$ where $Z$ is standard normal.

Compute. If the resulting probability is $p$, then our required probability is approximately $1-p$.

Remark: It is unfortunately quite possible to make an "off by $1$" error when using the continuity correction. To avoid that, it is useful to reduce our problem always to one of the shape $\Pr(X\le n)$, where $n$ is an integer, and to remember that this is approximately the probability that the approximating normal is $\le n+\frac{1}{2}$.

Alternately, if we are using the normal $W$ to approximate a random variable $X$ that only takes on integer values $k$, use the fact that $\Pr(X=k)\approx \Pr(k-\frac{1}{2}\le W\le k+\frac{1}{2}$.

$\endgroup$
  • $\begingroup$ Thanks André! This is much clearer now! $\endgroup$ – Ben G. Apr 29 '16 at 4:43
  • $\begingroup$ You are welcome. I will add a "practical" remark about the continuity correction. $\endgroup$ – André Nicolas Apr 29 '16 at 4:46
1
$\begingroup$

You have $\mu = \dfrac{100}{6}$ and $\sigma = \sqrt{\dfrac{500}{36}}$, and that is okay. $\checkmark$

You are thus looking for $\mathsf P\left(Z\gt \dfrac{15.5-\mu}{\sigma}\right)$, where $Z\sim\mathcal N(0,1^2)$, as your Normal Approximation value.

Plug in your values and use your Z-tail tableau, online calculator, mathematical package, or whatever.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.