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.
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