1. Random Functions

At a heat-treating company, iron castings and steel forgings are heat-treated to achieve desired mechanical properties and machinability. One steel forging is annealed to soften the part for each machining. Two lots of this part, made of 1020 steel, are heat-treated in two different furnaces. The specification for this part is 36-66 on the Rockwell G scale. Let X1 and X2 equal the respective hardness measurements for parts selected randomly from furnaces 1 and 2. Assume that the distributions of X1 and X2 are N(47.88, 2.19) and N(43.04, 14.89), respectively.

What are the p.d.f.s of X1 and X2?
Compute P(X1>X2), assuming independence of X1 and X2.

2. Originally Posted by calabrone
At a heat-treating company, iron castings and steel forgings are heat-treated to achieve desired mechanical properties and machinability. One steel forging is annealed to soften the part for each machining. Two lots of this part, made of 1020 steel, are heat-treated in two different furnaces. The specification for this part is 36-66 on the Rockwell G scale. Let X1 and X2 equal the respective hardness measurements for parts selected randomly from furnaces 1 and 2. Assume that the distributions of X1 and X2 are N(47.88, 2.19) and N(43.04, 14.89), respectively.

What are the p.d.f.s of X1 and X2?
Compute P(X1>X2), assuming independence of X1 and X2.

You are given that the distribution of X1 is N(47.88,2.19), that is it is normal with mean 47.88 and variance 2.19. Look up the pdf of the normal distribution.

Same procedure for X2.

X3 = X1-X2 ~N(47.88-43.04, 2.19+14.89)

So P(X1>X2)=P(X3>0).

CB

3. I'm not sure where to find the pdfs of the normal distribution.

Can I get the answer to the second part from the normal distribution table? I looked on the P(Z>z) table to find P(X3>0)=0.5000

Thank you

4. Originally Posted by calabrone
I'm not sure where to find the pdfs of the normal distribution.
wikipedia: normal distribution

Can I get the answer to the second part from the normal distribution table? I looked on the P(Z>z) table to find P(X3>0)=0.5000

Thank you
You compute the z-score corresponding to X3=0 when X3~N(47.88-43.04, 2.19+14.89), (and remember that the second argument here is the variance not the SD). Now look up the z-score in the normal table.

CB

CB

5. Would you be able to show me the work for this problem or one similar so I can follow your example to solve my remaining problems? Thank you for all your help.

6. Originally Posted by CaptainBlack
wikipedia: normal distribution

You compute the z-score corresponding to X3=0 when X3~N(47.88-43.04, 2.19+14.89), (and remember that the second argument here is the variance not the SD). Now look up the z-score in the normal table.
The z-score corresponding to $x_3=0$, when $X_3 \sim N(47.88-43.04, 2.19+14.89)=N(4.84,17.18)$ is:

$z=\frac{x_3-4.84}{\sqrt{17.18}}=\frac{0-4.84}{\sqrt{17.18}}=-1.16771$

Now we look $-1.16771$ up in the normal table or using a normal distribution calculator (available on line here) to get the required probability of $\approx 0.1215$

CB

7. Thank you so much. For the first part, I still do not know what chart I should get the pdfs from. I attached the file of my work showing a graph of the pdfs, but I don't know how to evaluate their equations to get a normal pdf. Thank you again.