# Question on normal distribution #2

• Sep 15th 2010, 09:51 PM
mngeow
Question on normal distribution #2
A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal.

What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold?

How should I go about doing this question?
• Sep 15th 2010, 10:00 PM
Unknown008
So, let X be the number of hours of life span of a bulb.

X ~ N(1200, 200^2)

P(X < x) = P(Z < z) = 0.05

Find the z value that corresponds to this probability, then convert it into hours using:

$\displaystyle z = \dfrac{x - \mu}{\sigma}$
• Sep 15th 2010, 10:25 PM
mngeow
I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it.
• Sep 15th 2010, 11:17 PM
MathoMan
Quote:

Originally Posted by mngeow
A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal.

What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold?

How should I go about doing this question?

X is normally distributed random variable $\displaystyle X\sim(1200,200^2)$ measuring the life span of light bulbs.

You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs.

$\displaystyle P(X>A)=0.95$
$\displaystyle 1-P(X<A)=0.95$
$\displaystyle P(X<A)=0.05$
$\displaystyle F_X(A)=0.05$
$\displaystyle \Phi(\frac{A-1200}{200})=0.05$

Now use the table values for standard normal distribution and you'll see that $\displaystyle \Phi(-1.65)\approx 0.05$ so you form the equation
$\displaystyle \frac{A-1200}{200}=-1.65$
$\displaystyle A=870.$
• Sep 16th 2010, 01:25 AM
mngeow
Quote:

Originally Posted by MathoMan
X is normally distributed random variable $\displaystyle X\sim(1200,200^2)$ measuring the life span of light bulbs.

You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs.

$\displaystyle P(X>A)=0.95$
$\displaystyle 1-P(X<A)=0.95$
$\displaystyle P(X<A)=0.05$
$\displaystyle F_X(A)=0.05$
$\displaystyle \Phi(\frac{A-1200}{200})=0.05$

Now use the table values for standard normal distribution and you'll see that $\displaystyle \Phi(-1.65)\approx 0.05$ so you form the equation
$\displaystyle \frac{A-1200}{200}=-1.65$
$\displaystyle A=870.$

I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that?
• Sep 16th 2010, 01:31 AM
Unknown008
Quote:

Originally Posted by mngeow
I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it.

That means that you are asked to use a z table different from mine.

So, from where I left:

$\displaystyle P(Z < z) = 0.05$

Is your graph like this: http://p1cture.me/images/51394697395543890592.png
Or like this: http://p1cture.me/images/88238005700832427228.png

?
• Sep 16th 2010, 01:34 AM
mngeow
Its the one in the 2nd link,where the shaded area is on the right hand side.
• Sep 16th 2010, 01:47 AM
Unknown008
Ok.

You need to find the z value for this area: http://p1cture.me/images/17640253637132137970.png

But you don't have this in your table. The equivalent of this area is this area: http://p1cture.me/images/50823189974773945850.png

However, this isn't in your table either. But you have this area: http://p1cture.me/images/30470114981255486093.png

The probability of the last area is 0.5 - 0.05 = 0.45, okay?

What is the value of z for this area? in your table, you should read z = 1.65.

But when you look at the original distribution (first graph), you know that z must be negative as it's value is on the left of the mean, so, the z value is -1.65.

From there, you can then work back the value of X, the maximum lifetime of the bulb for it to fall in the 5% worst quality.

Is it clear now?
• Sep 16th 2010, 01:52 AM
mngeow
Yes.I understand now.Thanks :D
• Sep 16th 2010, 05:07 AM
MathoMan
Quote:

Originally Posted by mngeow
I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that?

$\displaystyle F_X(A)$ is the value of the distribution function of the random variable X at point A.
$\displaystyle \Phi$ a uppercase Greek letter phi, commonly used to denote the distribution function of the standard normal random variable.

If X is normal random variable with parameters called expectation and variance, $\displaystyle X\sim {\cal N}(\mu, \sigma ^2)$ then distribution function of that random variable is usually denoted with $\displaystyle F_X$ and it is defined as $\displaystyle F_X(a)=P(X<a)$
Standard normal random variable is normal random variable with parameters 0 and 1, and to distinct it from other normal variables its distribution function, although defined in the same manner, is usually denoted with $\displaystyle \Phi$:
$\displaystyle X\sim {\cal N}(0, 1) , F_X(a)=\Phi(a)=P(X<a)$