Which is the probability that a group of 10 people exceed the maximum load? 
An elevator with capacity for 10 people is designed to support a maximum load of 750kg.If the weights  of people that use the elevator are distribuited with mean 70kg and standard deviation 8kg. Which is the probability that a group of 10 people exceed the maximum load ? 

Attempt: 
I know that I should have something like this $P(X>750)=1-P(X\le 750)=..$ 
but I don't know how to define the random variable $X$. I also don't know what is the distribution.
I was thinking on 'Let $X$ be the random variable that represent the weights of 10 people'.
Am I correct?
Could someone guide me please?
 A: If you are assuming a normal distribution then $X_i \sim \mathcal{N}(70, 8^2)$. Therefore,
$$
X=\sum_{i=1}^{10} X_i \sim \mathcal{N}(700, 640)
$$
Finally,
$$
P(X\leq 750) = P\left(\frac{X-700}{\sqrt{640}} \leq \frac{50}{\sqrt{640}}\right) = P\left(Z \leq \frac{50}{\sqrt{640}}\right)
$$
where $Z$ is the Standard normal. So you can find the value for $P(Z\leq x)$ is a table. We should obtain a probability of approximately 98%. That is, $P(X>750) \approx 0.02$.
A: The total weight $T$ of ten such people is distributed as 
$\mathsf{Norm}(\mu = 700,\, \sigma = \sqrt{640} = 25.2982).$
You seek $P(X > 750) = 1 - P(X \le 750) = 0.0241.$ You can evaluate this
probability using software (I used R statistical software below) or
by standardizing and using printed tables of normal probabilities. 
1 - pnorm(750, 700, 25.2982)
## 0.02405332

A: When adding independent normally distributed random vars, their sum is normally distributed where the mean is the sum of the means and the variance (square of standard deviation) is the sum of the variances.
So here we get that the mean is $10\cdot 70\,\rm{kg}$ and the variance is $10\cdot 64\,\rm{kg}^2$
A: The answers here love to indulge in normal distributions but that is a red herring.  Obviously weight is not normally distributed since the probability of one person weighing more than double of another is non-zero while the probabily of one person weighing negative (again, a deviation of more than the other person's weight from the other person's weight, just downwards instead of upwards) is zero.
However, a reasonable assumption here is that the various weights can be considered independent, with the value one person weighs not adding more information with regard to the value of another than what our starting information about the distribution tells us.
When adding independent random variables, the cumulants add.  The first cumulant is the mean, the second cumulant is the variance, with standard deviation usually defined as the square root of the variance.
Now it turns out that the normal distribution is the least compact distribution (has the largest area in its tails given a particular variance), so we can do the calculation of the resulting distribution as if it were a normal distribution, and the answer will take the form
"the probability that a group of 10 people will exceed the load will be at most ..." given the stated conditions where the estimate will be exact for the asymptotically often tenable stipulation of purely normal distribution (because the probability of negative weights is small enough to disregard).
A: Figure out the Z score for a mean of 75 kg in a normal distribution with a mean of 70 kg and SD of 8 kg. Z = (75 - 70)/(8/sqrt 10). From a Z table read off the p value and subtract it from 1, which is then the area to right of the Z score as the probability of exceeding the max weight. You can also do this on a TI-83 or similar as test 1 (Stat, Test, Z-test). Input the data and choose u > uo.
