Central Limit Theorem Problem with Uniform Distribution Just want to know if someone can check my work on this.
Individual claim sizes may be modeled by a uniform distribution function on the interval $[1000, 5000]$. Eighty claims are examined. Assume claims are independent.
What is the approximate probability that total claims will be between $250,000$ and $300,000$?
Here's what I did: $$\mu = {(a+b)\over2}=3000$$
$$\sigma^2={(b-a)^2\over12} = {4000^2\over12}={16000000\over12}$$
$$\sigma = \sqrt{\sigma^2}=\sqrt{16000000\over12}=1154.7$$
New $\mu = 3000*80 = 240,000$ and new $\sigma = 1154.7*80=92376.04307$
So we want $$Pr({10000\over92376.04307}<z<{60000\over92376.04307})$$
$$=P(z<.6495)-1+P(z<.108)$$ I end up getting a negative number which isn't right, what am I missing?
 A: You have a sum of 80 independent random variables. Let´s say $Y=\sum\limits_{i=1}^{80}X_i$, where $X_i\sim U(1000,5000)$
Then $\mu_Y=80\cdot 3000$. And the variance of the sum of independent random variables is equal to the sum of the  variances.
$Var\left(\sum\limits_{i=1}^n X_i\right)=\sum_{i=1}^n Var(X_i)$
Therefore $Var(Y)=80\cdot \frac{4000^2}{12}$
Applying central limit theorem
$$P(250,000\leq Y\leq300,000)\approx\Phi \left( \frac{300,000-240,000}{\sqrt{80\cdot \frac{4000^2}{12}}}\right)-\Phi \left( \frac{250,000-240,000}{\sqrt{80\cdot \frac{4000^2}{12}}} \right)$$ $$=\Phi(5.81)-\Phi(0.968)=1-0.8334=0.1666=16.66\%$$
A: Lets look at the exact form of the central limit theorem given by LL
$$\sqrt{n}(\bar{X}-\mu)\rightarrow^d N(0,\sigma^2)$$
where $\mu=E(X_i)$ and $\sigma^2=Var(X_i)$
So you did your calculations correctly above for the individual $X_i$, therefore we get
$$\sqrt{80}(\bar{X}-3000)\rightarrow^d N(0,1154.7^2)$$
Now we are interested in the total not $\bar{X}$ so we can multiply by $n$ on both sides and we get 
$$\sqrt{80}(\sum X_i-80*3000)\rightarrow^d N(0,80^2*1154.7^2)$$
which gives us that 
$$\sqrt{80}(\sum Xi-80*3000)\rightarrow^d N(0,80^2*1154.7^2)$$
$$\frac{\sqrt{80}(\sum Xi-80*3000)}{80*1154.7}\rightarrow^d N(0,1)$$
$$\frac{(\sum Xi-80*3000)}{\sqrt{80}1154.7}\rightarrow^d N(0,1)$$
Let $Y=\sum X_i$
$$\frac{(Y-80*3000)}{\sqrt{80}1154.7}\rightarrow^d N(0,1)$$
