2
$\begingroup$

Stones of a particular kind weigh in mean 10 kg and have standard deviation 1 kg. Assume the weight of a truck is normal distributed with mean 1000 kg and standard deviation 100 kg. 50 stones are put into a truck. Determine the probability that the total weight of the 50 stones and the truck is more than 1800 kg. Motivate your calculations.

what I did: $$ M = 1000 + (10*50) = 1500 $$ $$ \sigma = 100 + (1*50) = 150 $$ $$ P(X > 1800)=1-G(\frac{1800-1500}{150}) $$ $$ =1 - G(2) $$ $$ =1 - (erf(2) + \frac12) $$ $$ =\frac12 - (erf(2)) $$ $$ =\frac12 - .47726 $$ $$ =.02274 $$

Correct answer is $0.0014$. Where did I go wrong?

$\endgroup$
  • 1
    $\begingroup$ You cannot add the $\sigma$'s, I think, but their squares. $\endgroup$ – Henno Brandsma Nov 26 '14 at 20:27
2
$\begingroup$

The standard deviation calculation is not right.

Let $X$ be the weight of the truck and $Y$ the combined weights of the stones. The variance of $X$ is $10000$ and the variance of $Y$ is $(50)(1^2)$. So the variance of $X+Y$ is $10050$. So the standard deviation of $X+Y$ is $\sqrt{10050}$.

We have assumed that $X$ and $Y$ are independent. We have also assumed that the weights $W_1$ to $W_{50}$ of the $50$ stones are independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.