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Every time you go to a beach for vacation, you take home a little sand to keep as a souvenir. Over your lifetime, you have done this exactly 100 times. On each visit, the weight of sand you take home (measured in ounces) varies, but follows a $\mathrm{U}[10-\sqrt{48}/2, 10+\sqrt{48}/2]$ distribution. Assuming the amounts you take home on each trip are independent, estimate the probability that you have collected at least $980$, but no more than $1030$ ounces of sand.

I began by assigning $X$ as the total ounces of sand where $P(980 \le X \le 1030)$ is the probability we want to find. Then I set $Y$ as the ounces of sand per trip so that the sum of $Y$'s would equal $X$. From there, do I find the probability of $Y_i$, the ounces of sand during trip $i$, using uniform distribution? Then do I raise it to the power of 100 to find $X$?

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    $\begingroup$ The series of independent uniformly distributed random variables is approximately normally distributed for a large sequence. $\endgroup$ Jul 24 '14 at 23:24
  • $\begingroup$ @GrahamKemp So would I find the mean (expected value) and variance of Y and multiply it by 100 to find the probability for X? $\endgroup$
    – Eric
    Jul 24 '14 at 23:27
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Outline: Let $Y_i$ be the amount of sand brought back on trip $i$. Then $$X=Y_1+Y_2+Y_3+\cdots +Y_n,$$ where $n=100$.

The mean $\mu$ of each $Y_i$, by symmetry, is easy to find. The variance $\sigma^2$ is less easy. You can look it up, or calculate it from first principles. By the way, if $W$ is uniformly distributed on the interval $[a,b]$, then $W$ has variance $\frac{1}{12}(b-a)^2$. So the variance turns out to be quite nice in our case.

The exact distribution of $X$ is complicated. However, the $Y_i$ are nicely behaved independent identically distributed random variables, and $n$ is fairly large.

Thus (Central Limit Theorem) $X$ has approximately normal distribution, mean $100\mu$, variance $100\sigma^2$.

Now the probability that $X$ is between $980$ and $1030$ can be estimated.

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Lindeberg–Lévy Central Limit Theorem.

Suppose $\{X_1, X_2, ..., X_n\}$ is a sequence of i.i.d. random variables with $E[X_i] = \mu$ and $Var[X_i] = \sigma^2 < \infty$. Then as $n$ approaches infinity, the random variables $\sqrt{n}(S_n − \mu)$ converge in distribution to a normal ${\cal N}(0, \sigma^2)$

$$\lim_{n\to\infty}\sqrt{n}\bigg(\mathop{\frac{1}{n}\underbrace{\bigg(\sum_{i=1}^n X_i\bigg)}}_{\text{You wish to find this.}} - \mu\bigg) \sim {\cal N}(0, \sigma^2)$$

That is: $\sum\limits_{i=1}^n X_i \,\dot\sim\,{\cal N}(n\mu, n\sigma^2)$

So then, you need to find the mean ($\mu$) and variance ($\sigma^2$) of an individual sample, when: $$X_i\sim{\cal U}(10-\sqrt{12}, 10+\sqrt{12})$$

Find: $\Pr(\frac{980-n\mu}{\sqrt{n\sigma^2}}\leq Z \leq \frac{1030-n\mu}{\sqrt{n\sigma^2}})$ using your Z-Tables.

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