Problem about normal distribution I have the following problem:
A truck transports oranges. Each orange has a mean weight of 85g and a sd of 9 g.
If the truck transports 2000 oranges what is the probability that the total weight is greater than 175kg?
Now suppose that there are 3400 oranges and the total weight is 255kg. What is the probability that the mean weight of each orange is greater than 76g?
My attempt to the solution for the first question is obvious:
$\sigma_{\bar{X}}=9/\sqrt{2000}$ and $\bar{X}=175000/2000=87.5g$. So I have to find
\begin{equation}
P\left(\frac{87.5-85}{\sigma_{\bar{X}}}<Z\right)=P(12.5<Z)
\end{equation}
Which is essentially zero. This defies my intuition somehow... Am I doing something wrong?
For the second question I am a bit stuck. In this case $\bar{X}=255000/3400=75g$ and $\sigma_{\bar{X}} = 9/\sqrt{3400}$. I think that the probability that the mean is greater than 76g is the same as saying what is the probability $P(Z<-6.47)$ but again this is counterintuitive... What am I doing wrong?
 A: I don't see why you need any statistical calculations for the second problem.
There are 3400 oranges with a total weight is 255,000 g. So the average weight of an orange is 255000/3400= 75 grams. That is exact. There is zero (0.000000000000....) chance that the average weight is greater than 76 grams. The average weight of oranges in that truck is 75 grams exactly. 
A: If we assume that the weight of an orange is normally distributed and that the weights of oranges are independent. Let $X$ be the weight of each orange. So:
$$X\sim N(\mu,\sigma^{2})$$
Define $Y$ as weight of oranges in the truck:
$$Y=\sum_{i=1}^{2000}X\sim N(2000\mu,2000\sigma^{2})$$
Therefore you want:
$$\begin{align}
P\big(Y>175\big)&=P\Bigg(Z>\frac{175-2000(0.085)}{\sqrt{2000(0.000081)}}\Bigg)\\
&=P(Z>12.4226)\\
&=1-P(Z<12.4226)\\
&\approx 0
\end{align}$$
For the second question, we define:
$$\begin{align}
\bar{X}&=\dfrac{1}{3400}\sum_{i=1}^{3400}X_{i}=\dfrac{255000}{3400}=75
\end{align}$$
where $X_{i}$ is the weight of the $i$th orange.
We know the standard deviation of $\bar{X}$ is:
$$\text{SD}_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{9}{\sqrt{3400}}$$
Finally, we want to know:
$$\begin{align}
P\big(\mu>76\big)&=P\Bigg(Z>\frac{76-75}{9/\sqrt{3400}}\Bigg)\\
&=1-P\Bigg(Z<\frac{76-75}{9/\sqrt{3400}}\Bigg)\\
&\approx 0
\end{align}$$
