Estimating Mean Weight of Turkeys: Calculating Probability of Error

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Weight of turkeys is normally distributed with a standard deviation of 9 pounds. Farmer Jones samples 25 turkeys from his farm in order to estimate their mean weight. What is the probability that his sampling error will not exceed 2 pounds?

I hate this because I can't figure out what formula to use? I found one for the standard error of the mean where

SE = \frac{s}{\sqrt{n}}

This gives me the answer of 1.8. Now if I figure out the probability of exceeding 2 lbs how do I determine this with my given answer?
 
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You know the weights are normally distributed, correct? You know the value of \sigma but not \mu, so using the normal distribution is out of the question.

Here's a hint masquerading as a question: what do you know about the distribution of the sample variance when you sample from a normal distribution?
 
I am not sure if I am answering exactly what you are looking for, but if you know the standard deviation you will know the standard variance since it is sd squared. What this exactly tells me about the distribution I am unsure.
 
Ok...sorry, but I am still not understanding what exactly I am supposed to see with this problem. Any help?
 
Any help, anybody? More than happy to figure it out if just given a push in right direction.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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