General Statistics Problem

[SSmith]

Heres the problem:

5. A dimension under control is assumed to be normally distributed with a mean of 6.55 and a standard deviation of 0.15. The acceptable range for this dimension is 6.45-6.80. The percent of product that is nonconforming is most nearly:

A. 4.7%

B. 24.4%

C. 25.2%

D. 29.9%

Heres how Im working the problem:

P(X<6.45)+P(X>6.80)

P(Z<((6.45-6.55)/.15)+P(Z>((6.80-6.55)/.15)

P(Z<-0.67)+P(Z>1.67)

0.2743+0.0548

0.3291

Heres how the solutions says to work the problem:

P(X<6.45)+P(X>6.80)

P(Z<((6.45-6.55)/.15)+P(Z>((6.80-6.55)/.15)

P(Z<-0.67)+P(Z>1.67)

(0.2743-0.7x0.0323)+(0.0548-0.007)

0.299

If anyone can help explain where the red section comes from, I would appreciate it. This is a problem listed in the FE Industrial Discipline Sample Questions so its not helping my confidence level about Friday's PE. *sighs*

 

[benbo]

Heres the problem:
5. A dimension under control is assumed to be normally distributed with a mean of 6.55 and a standard deviation of 0.15. The acceptable range for this dimension is 6.45-6.80. The percent of product that is nonconforming is most nearly:

A. 4.7%

B. 24.4%

C. 25.2%

D. 29.9%

Heres how Im working the problem:

P(X<6.45)+P(X>6.80)

P(Z<((6.45-6.55)/.15)+P(Z>((6.80-6.55)/.15)

P(Z<-0.67)+P(Z>1.67)

0.2743+0.0548

0.3291

Heres how the solutions says to work the problem:

P(X<6.45)+P(X>6.80)

P(Z<((6.45-6.55)/.15)+P(Z>((6.80-6.55)/.15)

P(Z<-0.67)+P(Z>1.67)

(0.2743-0.7x0.0323)+(0.0548-0.007)

0.299

If anyone can help explain where the red section comes from, I would appreciate it. This is a problem listed in the FE Industrial Discipline Sample Questions so its not helping my confidence level about Friday's PE. *sighs*

I can't really help, but I have a guess.

I'm just guessing here, because I didn't really study this in engineering - I had a class with control limits in business school. So I'm really rusty on standard deviation curves. Did you look these probabilites up in a table? Is it possible these could be interpolation calculations, or did you hit these numbers right on the money?

I wouldn't worry too much. Anyway, the good news is that with your calculations you would have still picked the right answer.

Hopefully some six-sigma black belt will answer.

 

[SSmith]

I can't really help, but I have a guess.I'm just guessing here, because I didn't really study this in engineering - I had a class with control limits in business school. So I'm really rusty on standard deviation curves. Did you look these probabilites up in a table? Is it possible these could be interpolation calculations, or did you hit these numbers right on the money?

I wouldn't worry too much. Anyway, the good news is that with your calculations you would have still picked the right answer.

Hopefully some six-sigma black belt will answer.

Thats exactly what I was missing. Man am I out of practice with those tables. Im glad that I caught it now though because it hasnt come up with any of my other practice problems. I guess the texts Im using fit the problem to line up exactly with the tables they provide. Thanks for the help! (Sadly Im a six-sigma black belt also. Just have gotten used to Excel regurgitating what the value is instead of looking it up in a table.)

 

[Guest]

When I worked this problem, I got the same answer up until the point you did:

P(Z<-0.67) and P(Z>1.667)

Looking at the z-score tables, P(Z<-0.67) = 0.2514 and P(Z>1.667) = 0.9525.

When you resolved the probability to the z-score - this is giving you the % at face value. Based on your problem statement, you are looking for the probability for the areas that are at either tail.

So, the first tail simply = 0.2514 (based on knowing P(Z1)). In order to find the second tail, you use your knowledge that the total probability = 1; therefore, P(Z2) = 1 - 0.9525 = 0.0475

When you add those two numbers together, you get 0.2989 (0.299).

I hope this helps! Good luck and don't fret before taking the exam!

Regards,

JR