1. ## Normal distribution help

Q: To avoid a complete shutdown of the assembly line if a component fails, major manufacturing companies operate on the principle of preventative maintenance. the lifetime of one component is normally distributed with a mean of 325h and a standard deviation of 20h. How frequently should the component be replaced so that the probability of its failing during operation is less than 0.01?

A: I don't exactly understand it.

What exactly do I do? I tried stuff like P(x<0.01) etc but that doesnt work.

Is it something to do with making it =0.01?

2. Originally Posted by Johnboe
Q: To avoid a complete shutdown of the assembly line if a component fails, major manufacturing companies operate on the principle of preventative maintenance. the lifetime of one component is normally distributed with a mean of 325h and a standard deviation of 20h. How frequently should the component be replaced so that the probability of its failing during operation is less than 0.01?

A: I don't exactly understand it.

What exactly do I do? I tried stuff like P(x<0.01) etc but that doesnt work.

Is it something to do with making it =0.01?
Did you try to solve for $\displaystyle a$ where

$\displaystyle P\left( Z<\frac{a-325}{20}\right)= 0.01$

3. Originally Posted by pickslides
Did you try to solve for $\displaystyle a$ where

$\displaystyle P\left( Z<\frac{a-325}{20}\right)= 0.01$
so a=325.2 if I am correct?

No, grr, I don't get it.

4. Originally Posted by Johnboe
so a=325.2 if I am correct?

No, grr, I don't get it.
Get the z-value z* such that Pr(Z < z*)= 0.01. Now solve z* = (a - 325)/20 for a.

5. Originally Posted by Johnboe
so a=325.2 if I am correct?
Think about the curve itself and the case where $\displaystyle a = 325$

It gives

$\displaystyle P\left( Z<\frac{325-325}{20}\right)= P(Z<0 )=0.5$

therefore your answer needs to be much less than $\displaystyle 325$

6. z*=0.01 @ 0.5040

0.5040 = a-325 / 20

10.08+325=a
a=335.08

mm, still don't get it

7. This is the problem.

Originally Posted by Johnboe
z*=0.01 @ 0.5040
$\displaystyle z^{*} =0.01$ at $\displaystyle -2.3263$

Look here Z table - Normal Distribution

8. Where did you get -2.3263 from?

9. Originally Posted by Johnboe
Where did you get -2.3263 from?
Originally Posted by pickslides

10. Yes, I looked there... but it just has 2 graphs and I played around and I do not get -2.3263 anywhere.

11. Originally Posted by Johnboe
Yes, I looked there... but it just has 2 graphs and I played around and I do not get -2.3263 anywhere.

Go to Z table - Normal Distribution

Under the second curve, under "Shadded Area" put 0.01 and tick "Below" A figure will pop up next to "Below"

12. Which then solves to a=278.474?

13. Which means: the component should be replaced every 278.5 hours so it fails less than 1%?

14. Thats how I read it.