# Approximating Chi squared distribution

A machine in a heavy equipment-factory produces steel rods of length Y , where Y is a normally distributed random variable with mean 6cm inches and variance $\frac{1}{4} cm^2$. Thecost C of repairing a rod that is not exactly 6 inches is given by$C = 4(Y - 6)^2$ . If 50 rods with independent lengths are produced in a given day, approximate the probability that the total cost for repairs for that day exceeds 48.

Alright, first things first: C looks like $\chi^2(1)$ because that 6 and 4 are not there without a reason: \begin{align} C &= 4(Y-6)^2 & \\ &=\frac{(Y-\mu)^2}{\sigma^2} &\mu=6, \sigma^2 = \frac{1}{4} \end{align} so $C\sim\chi^2(1)$, because $Y\sim N(\mu,\sigma^2)$ So we can assume that $50* \chi^2(1) = \chi^2(50)$ (is that a good way of saying that? )

So lets say $X = \chi^2(50)$ Our new question is: (after this point im lost) \begin{align} P(X>48) &= P(\frac{X-50}{100} >\frac{48-50}{100}) \\&\approx P(Z>-0.02) = P(Z<0.02) \\&\approx 0.5080 \end{align} Does this seem a reasonable way of doing things? I have no way of checking

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"with mean 6cm inches" cm-inches is a new one on me. Please make your use of units consistent. "$50 \chi^2(1) = \chi^2(50)$" -- this statement is false. Are you confusing adding 50 random variables with multiplying one random variable by 50? –  Glen_b Apr 24 '13 at 5:03
actualyl yeah, that was wrong in the exercise. And you are right i am confusing that. –  WiseStrawberry Apr 24 '13 at 9:44

If $Y_i$ are iid $N(\mu, \sigma^2)$, then $$\sum_{i=1}^{50}\left(\frac{Y_i-\mu}{\sigma}\right)^2\sim \chi^2(50)$$ as you indicated. Remember that the sum of $n$ squared independent standard normal variables is $\chi^2$ distributed with $n$ degrees of freedom. So you don't need to go via $C\sim \chi^2(1)$, you may go there immediately.
As this is a sum of independently and identically distributed random variables and $n$ is reasonably 'large', we may use the Central Limit Theorem for a normal approximation, as you also have done. However, you need to remember that the standardization is done in the following way:
$$Pr(X < c)=Pr\left(\frac{X-\mu_x}{\sigma_x}<\frac{c-\mu_x}{\sigma_x}\right)=Pr\left(Z<\frac{c-\mu_x}{\sigma_x}\right)$$ where we use the standard deviation $\sigma$. Note that the variance of a $\chi^2(v)$ is $2v$, meaning that the standard deviation must be $\sigma=\sqrt{2v}$. So you are almost there, you just have to cross the line, so to speak.
So my only fault was doing $\sigma = 2v$ instead of $\sigma \sqrt{2v}$ –  WiseStrawberry Apr 24 '13 at 9:46