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Under the assumption that a pair of dice is fair, the probability is 0.88 that the number of 9's appearing in 288 throws of the dice will lie within 32 $\pm$ K. What is K?

I have found the standard deviation using

$\sqrt{mean\:\cdot \:trials\:\cdot \:possibility\:of\:pair\:of\:fair\:dice\:show\:9}$

but I cannot go any further from that.

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    Not following your formula. For the standard normal approximation you'd get $\sigma =\sqrt {Np(1-p)}$ where $N$ is the number of trials and $p$ is the probability of success in a single trial. – 2017-01-18
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    @lulu Oh, I must have misunderstood something on the other similar examples. I take it I have to use the z-table now and multiply the standard deviation with the z value to find the answer. But I couldn't understand which value I should be looking for on the z-table. – 2017-01-18
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    It's not too hard to do it exactly. I'll post both the approximate and the exact methods below. – 2017-01-18

2 Answers 2

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We note that this is a standard Binomial distribution with probability $p=\frac 19$.

If we like, we can approximate this with a normal distribution..with mean $\mu=32$ and $\sigma =\sqrt {288\times \frac 19\times 89}=\frac {16}3$

We then seek $k$ such that $$\Phi(32+K,\mu,\sigma)-\Phi(32-K,\mu,\sigma)=.88$$

Here $\Phi$ denotes the cumulative distribution function for the associated normal distribution. That's easy enough to solve numerically as it is, but if you want to use standard functions it helps to remark that $$\Phi(32+K,\mu,\sigma)+\Phi(32-K,\mu,\sigma)=1$$ From which we conclude that $$2\Phi(32+K,\mu,\sigma)-1=.88\quad \implies \quad \Phi(32+K,\mu,\sigma)=.94$$

Which gives $K=8.292125838$.

Of course, $K$ is meant to be an integer for the true (discrete) situation so one is lead to suspect that $K=8$.

Of course, it is not difficult to do this exactly using the Binomial Distribution. Let $p_n$ denote the probability of getting exactly $n$ $9's$. We have $$p_n=\binom {288}n \left(\frac 19\right)^n\left(\frac 89\right)^{288-n}$$

It's not hard to compute these with mechanical assistance. Inspired by the normal approximation we start with $n=24$ and compute $$p_{24}=0.025039361,p_{25}=0.033051957,\cdots,p_{39}=0.030441785,p_{40}=0.023687514$$ We add these up to get $0.89005738$. So, I'd have liked $.89$ instead of $.88$ in the question, but $K=8$ appears to get the job done. if you only add from $p_{25}$ to $p_{39}$ you get $0.841330505$ which is too low

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Hint $\Pr$(pair of fair dice show 9) = $4/36 = 1/9$

$(1/9 + 8/9)^{288}$ models the outcomes of 288 trials

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    I have found the standard deviation using √mean⋅trials⋅possibilityofpairoffairdiceshow9 but I cannot go any further from that. Could you explain how I could go on? @M.Diggerson – 2017-01-18
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    @BerkeJimmyBakkaloğlu I'm afraid I'm just a recent high school graduate (about to start university and study maths or stats later on) so I don't know the exact sequence of steps, but the furthest I reached was P(n 9s in 288|n=32+/-K) = $\sum{r=-K}{K}\binom{288}{32+r}(1/9)^{32+r}(8/9)^{288-(32+r}$ = 0.88 which I think can be done using the probability distribution curve/integral calculus (from what little further reading I've done); hope some of this helps – 2017-01-18