
how would I do this.....
The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day? :confused:

Well, I would use this general result on 'combining errors' in things that are normally distributed:
If you have a variable $\displaystyle Z$ normally distributed about a mean value $\displaystyle \bar{Z}$ with variance $\displaystyle \sigma_{Z}^2$, and a function $\displaystyle F(Z)$, the function will be approximately normally distributed about $\displaystyle \bar{F} = F(\bar{Z})$with a variance
$\displaystyle \sigma_A^2 = \left( \frac{\partial F}{\partial Z} \right)^2 \sigma_Z^2 $
In this case, the variable is the length $\displaystyle L$ with $\displaystyle \bar{L} = 10cm, \sigma_L^2 = (0.01cm)^2.$
The function is the period $\displaystyle T(L)$ given by:
$\displaystyle T = 2\pi \sqrt{\frac{L}{g}}$
so $\displaystyle \frac{\partial T}{\partial L} = \pi \sqrt{\frac{1}{gL}}.$
You need to find the standard deviation $\displaystyle \sigma_T$ where $\displaystyle \sigma_T^2 = \left( \frac{\partial T}{\partial L} \right)^2 \sigma_L^2 = \pi^2 \frac{1}{gL} \sigma_L^2$.
Use $\displaystyle L = \bar{L}$ in that expression.
...and then, using the properties of the normal distribution, you can find the probability of $\displaystyle T$ being wrong by a certain amount (enough to make the clock slow down by a minute a day). There's quite a bit of calculating left for you to do, but hopefully that should help you on your way.
  
You may have a different way to do it... it really depends on what you've been taught. :)