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The exercise reads:

In the storage of a stock of $100Kg$ flour bags, a random error $X$ is committed whose density function is of the form $f(x)=k(1-x^2)$, if $-1<x<1$ and $f(x)=0$, otherwise.

  • $a)$ Calculate the probability that a sack of flour will pass from $99,5Kg$.
  • $b)$ What percentage of sacks will have a weight between $99,8$ and $100,2Kg$?

My question is, the relationship between the error and the $100Kg$. Are we talking about a margin of error of $+1 Kg$ and $-1Kg$ respectively? That is:

In the part $ a) $ I must calculate $ P (X< -0.5) $ and in the part $ b) $ $ P (-0.2 <X <0.2) $?

Thank you very much.

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As far as i can tell your thoughts are correct. As you said all you have to do is to calculate those integrals you are talking about. But be careful about (a). you have to calculate the probability of X being such as

100Kg+X > 99.5Kg that is X>-0.5Kg

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  • $\begingroup$ In this case it would be $ 1-P (X <-0.5) $? $\endgroup$ – emi Feb 13 '17 at 23:14
  • $\begingroup$ @AccomDuter yes $\endgroup$ – johny Feb 13 '17 at 23:16
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The exercise as stated is potentially ambiguous, but I believe that yours is the most plausible interpretation.

The only other possible interpretation I can think of is that $X$ is the percentage error, so that the final weight would be $100 (1+X)$. But, frankly, this would be quite a stretch.

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  • $\begingroup$ Stretch like this? $\endgroup$ – emi Feb 13 '17 at 23:01

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