# normal random variable distribution

i have such problem in the book of Applied statistic and probability for Enigneering and need some help to solve it.problem is following:

Let random variable X denote a measurement from a manufactured product,suppose target value for this measurement is m,for example ,X could denoted dimensional length and target might be 10 millimeters,The quality loss of process producing the product is defined to be expected value of $(k(X-m)^2)$ $,where k is constant ,that relates a deviation from target to loss measured in dollars. we have additional two situation 1.suppose that X is a continuous random variable with$E(X)=m$and$V(x)=\sigma^2$whit is quality loss of the process? 2.suppose that X is a continuous random variable with$E(X)=u$and$V(x)=\sigma^2$whit is quality loss of the process? here u denotes usual mean character(sorry i could not type exact form by Latex) because as it seems that,this is a normal distribution,we have formula which express probbaility density function of normal variable,no i did not understand how to use this formula for answer questions,also i have doubt, that does expected value means mean value?i mean if there is given that expected value is equal to$(k(X-m)^2)$,does it means that,instead of mean,i should put$(k(X-m)^2)$in the equation of probability density function of normal distribution?sorry if my question is too big,i could not describe it shortly.please help me - Hint:$E[aX] = aE[X], E[k(X-m)^2] = kE[(X-m)^2]$. Do you know the value of$E[(X-m)^2]$for a normal random variable with mean$m$and variance$\sigma^2$? How about for any other kind of random variable with mean$m$and variance$\sigma^2$? – Dilip Sarwate Apr 22 '12 at 12:20 no in book there is not given it – dato datuashvili Apr 22 '12 at 12:32 $E[X] = \mu $ will give you$E[X] = \mu $– Henry Apr 22 '12 at 12:58 Additional hint: First prove that$E[(X-m)^2] = E[(X-\mu)^2] + (m-\mu)^2$by expanding the terms in square brackets on both sides and using the linearity of expectation which for your case means$E[X^2+aX+b] = E[X^2]+aE[X]+b$. Expand out$(m-\mu)^2$. Now combine terms on the right and see if you have the same answer as on the left. Then, read the definition of the variance of$X$. Can you tell what is the value of$E[(X-\mu)^2]$? Since you know$m$and$\mu$, can you find the value of$E[(X-m)^2]$in terms of$m$,$\mu$, and$\sigma^2$? The value of$E[k(X-m)^2]=kE[(X-m)^2]\$? –  Dilip Sarwate Apr 22 '12 at 15:43