Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
"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
up vote 1 down vote accepted

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.

share|cite|improve this answer
So my only fault was doing $\sigma = 2v$ instead of $\sigma \sqrt{2v}$ – WiseStrawberry Apr 24 '13 at 9:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.