Normal distribution help

• Jan 11th 2010, 12:15 PM
Johnboe
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?
• Jan 11th 2010, 01:23 PM
pickslides
Quote:

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 $a$ where

$P\left( Z<\frac{a-325}{20}\right)= 0.01$
• Jan 11th 2010, 01:44 PM
Johnboe
Quote:

Originally Posted by pickslides
Did you try to solve for $a$ where

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

so a=325.2 if I am correct?

No, grr, I don't get it.
• Jan 11th 2010, 01:51 PM
mr fantastic
Quote:

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.
• Jan 11th 2010, 01:57 PM
pickslides
Quote:

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

Think about the curve itself and the case where $a = 325$

It gives

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

therefore your answer needs to be much less than $325$
• Jan 11th 2010, 02:11 PM
Johnboe
z*=0.01 @ 0.5040

0.5040 = a-325 / 20

10.08+325=a
a=335.08

mm, still don't get it
• Jan 11th 2010, 02:17 PM
pickslides
This is the problem.

Quote:

Originally Posted by Johnboe
z*=0.01 @ 0.5040

$z^{*} =0.01$ at $-2.3263$

Look here Z table - Normal Distribution
• Jan 11th 2010, 02:32 PM
Johnboe
Where did you get -2.3263 from?
• Jan 11th 2010, 02:39 PM
pickslides
Quote:

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

Quote:

Originally Posted by pickslides

(Rofl)
• Jan 11th 2010, 02:41 PM
Johnboe
Yes, I looked there... but it just has 2 graphs and I played around and I do not get -2.3263 anywhere.
• Jan 11th 2010, 02:47 PM
pickslides
Quote:

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"
• Jan 11th 2010, 02:52 PM
Johnboe
Which then solves to a=278.474?
• Jan 11th 2010, 02:54 PM
pickslides
(Rock)(Party)(Beer)
• Jan 11th 2010, 03:00 PM
Johnboe
Which means: the component should be replaced every 278.5 hours so it fails less than 1%?
• Jan 11th 2010, 03:02 PM
pickslides