# Estimated probability looks too large, what I am doing wrong?

A factory produces links for heavy metal chains. The research lab of the factory models the length (in cm) of a link by the random variable ${X}$, with expected value ${E(X) = 5}$ and variance ${Var(X) = 0.04}$. The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free.

Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains?

Number of links in a chain: ${n = 1002}$

Mean lengh of 1002 links: ${ \mu = 1002 \times 0.05 = 50.1 }$

We know variance, so we can find the standard deviation: ${ \sigma = \sqrt{0.04} = 0.2 }$

We are interested how often chain is shorter then 50 meters: ${ \bar{x} = 50 }$

Thus let define an appropriate test statistic: ${ \displaystyle z = \frac{ \mu - \bar{x} }{ \sigma } = \frac{50.1 - 50}{0.2} = \frac{0.1}{0.2} = 0.5 }$

Probability of getting value in a lower tail is: ${ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz }$

By making the substitution ${ z = \sqrt{2} x }$, we get: ${ \displaystyle \frac{1}{\sqrt{2\pi}} \int\limits_{0}^{0.5} e^{-\frac{z^2}{2}} dz = \frac{1}{\sqrt{\pi}} \int\limits_{0}^{\frac{0.5}{\sqrt{2}}} e^{-x^2} dx }$

As we can not solve integral above, let's consider ${ e^{-x^2} }$ as a Taylor series: ${ \displaystyle \sum\limits_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} }$

Then the approximate probability is: ${ \displaystyle P(z < 0.5) = \frac{1}{2} + \frac{1}{\sqrt{\pi}} \sum\limits_{n=0}^{\infty} \frac{(-1)^n (\frac{c}{\sqrt{2}})^{2n+1} }{ n! (2n+1) } \simeq 0.691462 }$

But this probability looks too large. What I am doing wrong?

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The random variable representing the length of the chain is $Y=\sum_{k=1}^{1002}X_k$, where $X_k$ gives the length of the $k$-th link. Variance is linear, so $\operatorname{Var}(Y)=1002\operatorname{Var}(X)=40.08$, and the standard deviation is about $6.33$ cm. An error of $-10$ cm corresponds to $z\approx-1.58$.

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But I am getting ${ z \simeq 0.0158 }$. It is positive and hundred times smaller. Could you please explain how you get ${ z \simeq -1.58 }$? –  Edward Ruchevits Nov 21 '12 at 22:39
It’s just $10/\sqrt{40.08}$; are you sure that you’re using consistent units? I’m using cm throughout. –  Brian M. Scott Nov 21 '12 at 22:40
But what is with minus sign? –  Edward Ruchevits Nov 21 '12 at 22:41
@Edward: Because $5000$ cm is below the mean of $5010$ cm. It’s a deviation of $-10$ cm. –  Brian M. Scott Nov 21 '12 at 22:44
OK, thank you very much! –  Edward Ruchevits Nov 21 '12 at 22:47
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You are using the wrong variance! The variance of the sum is the sum of the variances, by independence. So the variance is $(1002)(0.04)$. There is also a little issue of units. The mean is $5010$ cm. The standard deviation, when you compute it, will be in cm. One can convert this to metres, and use metres for the length of the chain. Or you can use cm for both.

Formally, if $X_1,X_2,\dots, X_n$ are the lengths of the links, and $X$ is the length of the chain, then $$X=X_1+X_2+\cdots+X_n.$$ By independence, the variance of the sum is the sum of the variances, so if $\sigma^2$ is the variance of any $X_i$, then $X$ has variance $n\sigma^2$.

As a further remark, the standard way to do the numerical work is not by a Taylor series calculation, but by using tables for the cumulative distribution of the standard normal (old tech) or by using appropriate software (new tech).

Thus you want to find the probability that a standard normal has value less than $$\frac{5000 -5010}{\sqrt{(1002)(0.04)}}.$$

So we want the probability that a standard normal is $\lt -1.57956$. Fom the usual tables, we can read off the approximate probability that $Z\lt 1.57956$. By symmetry, for our required probability we subtract this from $1$.

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Thank you for explanation! –  Edward Ruchevits Nov 21 '12 at 22:48