I am a bit confused about one of the exercises and haven't been able to find any clues on how to solve it from the book. Can somebody point me in the right direction? I've changed the problem a bit so I can use this example to practice and the problem from my book to solve myself.
There are X random numbers, independent from each other and they all have the uniform distribution on interval $(0,1)$ We look at the sample mean.
Approximate the probability $P (X<0.56)$ for $n=10$ Use the central limit theorem.
What I got so far: $p((X_1+X_2+...+X_{10})/10 < 0.56)$, $(X_1+X_2+...+X_{10})/10$ is the sample mean, which you can also write as $\bar{X}$, I will write it as bold for computer reasons for now.
$p(\bar{X} < 0.56)$
This results in $p(Z < 0.56- \mu/(\sigma/\sqrt{n}))$
Now for $\mu$, this is the mean. I'd say that this is $0.5$ as this is the usual estimator. As for $\sigma$, S is the usual estimator or the square root sample variance.
This is the formula I should be using. I know n=10 but I'm not sure on how to calculate the X and X
$(X_n-(X_1+X_2+...+X_{10})/10) (2/9)$ I don't have any results for $X_1$ to $X_{10}$, how can I calculate this? Is it possible to pick $X_1=0$ and $X_{10}=1$ and for $X_2$ to $X_9$ values in between?
Please help me to understand the problem!
The idea behind the central limit theorem is that when the number of observations in your sample is "large enough," the sample mean is approximately normally distributed. Specifically, if $$X_1, X_2, \ldots, X_n$$ are independent and identically distributed random variables, each with mean $\mu$ and standard deviation $\sigma$, then $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ is approximately normal with mean $$\operatorname{E}[\bar X] = \mu,$$ and variance $$\operatorname{Var}[\bar X] = \frac{\sigma^2}{n}.$$ Therefore, $$Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}} \sim \operatorname{Normal}(0,1);$$ that is to say, $$(\bar X - \mu)/(\sigma/\sqrt{n})$$ is approximately standard normal. This allows us to obtain approximate probabilities, regardless (for the most part) of the distribution of the random variables $X_1, \ldots, X_n$, for the behavior of the sample mean.
So in your case, $$\Pr[\bar X < 0.56] = \Pr\left[\frac{\bar X - \mu}{\sigma/\sqrt{n}} < \frac{0.56 - \mu}{\sigma/\sqrt{n}}\right] \approx \Pr\left[Z < \frac{0.56 - 0.5}{(1/\sqrt{12})/\sqrt{10}}\right] = \Pr[Z < 0.657267].$$ Here what I have done is used the fact that for a uniform distribution on $[0,1]$, $$\mu = \frac{1}{2}, \quad \sigma^2 = \frac{1}{12}.$$ Now all that remains is to use a standard normal distribution table to look up the left-tailed probability; you should find that the desired probability is approximately $0.744495$.
Note carefully where I have placed the approximate equality symbol $\approx$; this is where the CLT is applied.
It is also worth mentioning that the exact distribution of the sample total of continuous uniform random variables on $[0,1]$ is called the Irwin-Hall distribution, and it can be used to obtain the exact probability, which is $$\frac{4380855272971}{5906250000000} \approx 0.741732.$$