Probablity question with unknown distribution function. 
A manufacturer produces washers with a thickness $d$ that has a mean value of 1.0 mm and a standard deviation of 0.1 mm. If $N = 25$ washers are selected at random and stacked on top of one another, determine the probability that the height of the stack will be between 24 mm and 26 mm assuming that $d$ has an unknown distribution function.

Can I just use the fact that each washer is iid and thus use the $E[Nd] ~$ Normal($1.0N,0.1N$)?
That is the only way I could think of solving this problem, but I am uncertain because the washers thickness is not normal. 
 A: Please allow me to modify your question in terms of just the variables. In other words, let the number of washers selected be $n=25$. Since $n$ is sufficiently large, we can apply Central Limit Theorem. The distribution of the sample follows a normal distribution. Give this distribution a mean of $\mu_\bar{x}$ and a SD of $\sigma_\bar{x}$. Here's how to calculate the 2 parameters of this new normal distribution:
$$
\begin{align*}
\mu_\bar{x} &= \frac{\sum x_1 + x_2 + ... + x_{25}}{25} \\
&=\frac{25 \times1.0}{25}\\
&=1.0
\end{align*}
$$
Now, for the $\sigma$, it is a little different
$$
\begin{align*}
\sigma^2_\bar{x} &= \frac{\sigma_1}{25} \\
&=\frac{0.1}{25}\\
&=0.004
\end{align*}
$$
A: Let $H$ be the height of the stack, and let $X_1,X_2,\dots,X_{25}$ be the thicknesses of the washers, say from the bottom of the stack up. Then
$$H=X_1+X_2+\cdots+X_{25}.$$
So $H$ is a sum of $25$ independent identically distributed random variables.  If the distribution of the $X_i$ is not too unreasonable, the total height $H$ will be approximately normally distributed.
The mean of a sum is the sum of the means, so $E(H)=25$.
The variance of a sum of independent random variables is the sum of the variances. It follows that
$$\text{Var}(H)=(25)(0.1)^2=0.25.$$
It follows that $H$ has standard deviation $\sqrt{0.25}=0.5$.
Now we need to find the probability that a normally distributed random variable with mean $25$ and standard deviation $0.5$ lies between $24$ and $26$. Note that $\dfrac{H-25}{0.5}$ is standard normal. So we want 
$$\Pr\left(-\frac{1}{0.5} \le Z\le \frac{1}{0.5}\right),$$
where $Z$ is standard normal. We can get this information from tables, or software.
By the way, a table gives $\Pr(Z\le 2)\approx 0.9772$. From this you should be able to calculate the required probability.  
