A fair die is thrown 12,000 times. Use the central limit theorem to find values of $a$ and $b$ such that $$ \mathbb P(1900<S<2200)\approx\int_a^b\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,\mathrm dx, $$ where $S$ is the total number of sixes thrown.

Right, so the central limit theorem goes as follows:

Let $X_1,X_2,\dots$ be independent and identically distributed random variables, each with mean $\mu$ and non-zero variance $\sigma^2$. The standardised version $$ Z_n=\frac{S_n-n\mu}{\sigma\sqrt{n}} $$ of the sum $S_n=X_1+\dots+X_n$ satisfies, as $n\to\infty$ $$ \mathbb P(Z_n\leq x)\to\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}u^2}\,\mathrm du\quad\text{for }x\in\mathbb R. $$

Okay, so it's obvious that I need to express $Z_n$ in terms of known quantities. For $S<2200$, we basically want $$ Z_n<\frac{2200-12,000\cdot12,000\cdot\frac{1}{6}}{\sqrt{12,000\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot12,000}}, $$ where I used $\mu=np$ and $\sigma^2=np(1-p)$, where $p=\frac{1}{6}$.

This approach gives me a numerical value of -5366.07.

I'm doing something wrong here. Could someone point out my mistake?


OK, I got the solution.

  • 1
    $\begingroup$ Think about what the $X_i$ are precisely in your example, and check their means and variances. $\endgroup$ – Matthew Towers Feb 2 '17 at 17:23
  • $\begingroup$ @m_t_ Oh right, I thought I had to choose $\mu$ and $\sigma^2$ for $S_n$, but I have to chose them for $X_i$, for each $i\in\mathbb N$. $\endgroup$ – Sha Vuklia Feb 2 '17 at 17:28
  • $\begingroup$ @m_t_ But then I get $$ Z_n<\frac{2200-12,000\cdot\frac{1}{6}}{\sqrt{\frac{1}{6}\cdot12,000}}=2\sqrt{5}, $$ while my solutions say it's $2\sqrt{6}$. Am I still doing something wrong? Edit; Oh, again a mistake. I thought the variance was $p$, but apparently it's $np$. $\endgroup$ – Sha Vuklia Feb 2 '17 at 17:40
  • $\begingroup$ The variance of $X_i$ isn't $p$ or $np$, it's $p(1-p)$. That will give you the answer you want. $\endgroup$ – Matthew Towers Feb 2 '17 at 19:42

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