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The amount,$A$ ,of wine in a bottle $~ N(1.05l,0.0004l^2)$ Casks are available which have a volume,$V$, which is $N(22l,0.16l^2)$.

What is the probability the contents of $20$ randomly chosen bottles will fit inside a randomly chosen cask?The answer is $0.9927$.

I do not know how to get the answer .Please help. Thank you.

Mhenni Benghorbal
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Jing
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  • How can $A$ be a normal variable . How can it take negative values ? – abkds Dec 04 '13 at 05:52
  • What is the sum of $20$ of the same normal distrubutions (hint: it is normal)? Don't we want to see the overlap of area of the probability densities of the 20 normals and the distribution of the cask normal. Does comparing two normal distribution sound like something familiar. –  Dec 04 '13 at 05:54
  • @TrafalgarLaw, I think the variance is low enough that the probability of anything negative is virtually 0. –  Dec 04 '13 at 05:55
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    Ah , I see it is an approximation ! – abkds Dec 04 '13 at 05:56

1 Answers1

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Let $X_1,X_2,\dots, X_{20}$ be the amounts of wine in the various bottles, and let $V$ be the volume of the cask. Let $$W=X_1+X_2+\cdots +X_{20}-V.$$ We want to find the probability that $W\le 0$.

We assume the $X_i$ and $V$ are independent. Then $W$ has normal distribution.

The mean of $W$ is $(20)(1.05)-22$, that is, $-1$.

The variance of $W$ is $(20)(0.0004)+0.16$. (For all practical purposes, it is determined by the variance of $V$.)

So we want to find the probability that a normal of known mean and variance is $\le -1$. This is done via a standard calculation, using software or a table of the standard normal.

André Nicolas
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