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?