# flaws in a tape

• Jul 6th 2011, 06:45 PM
wopashui
flaws in a tape
The number of flaws on a magnetic tape produced continuously at a factory follows a Poisson distribution with an average of 0.01 falws per meter. A stand unit of this tape contains 250 meters of magnetic tape.

(a) What is the probability that there are at least two flaws in a single unit of tape?

(b) In a random sample of 20 units of tapes, what is the probability that at least two of them are flawless?

for part a I have 250x0.01=2.5 P(x>=2)=1-P(x=0)-P(x=1)=1-e^{-2.5}*2.5^0/0! -e^-2.5*2.5^1/1!

for part b, do I just times 2.5 by 20?
• Jul 6th 2011, 07:27 PM
TKHunny
Re: flaws in a tape
You seem to have part (a). Good work.

Not so good on part (b). First, "times" isn't a verb.
Second, you now have a binomial problem.

n = 20

$p = \frac{e^{-2.5}\cdot 2.5^{0}}{0!}$
• Jul 6th 2011, 09:31 PM
wopashui
Re: flaws in a tape
Quote:

Originally Posted by TKHunny
You seem to have part (a). Good work.

Not so good on part (b). First, "times" isn't a verb.
Second, you now have a binomial problem.

n = 20

$p = \frac{e^{-2.5}\cdot 2.5^{0}}{0!}$

and what does this mean then, do i use binomial approximation to possion?
• Jul 6th 2011, 09:58 PM
pickslides
Re: flaws in a tape
Don't need the approximation. You just need n=20 from the question and p from part a)
• Jul 7th 2011, 04:03 AM
TKHunny
Re: flaws in a tape
Please stop guessing. Go reread the section on Binomial, Poisson, Bernoulli, and other discrete distributions.
• Jul 7th 2011, 08:57 AM
wopashui
Re: flaws in a tape
Quote:

Originally Posted by pickslides
Don't need the approximation. You just need n=20 from the question and p from part a)

no, since there is 0.01 flaws per meter, then 20 units has 20*250 meter, so n is 5000
• Jul 7th 2011, 09:43 AM
SpringFan25
Re: flaws in a tape
As previous posters have said, part (b) is a binomial question.

Step 1 Find probability that any given sample is flawless

The number of flaws in a sample of length 250 is $X \sim Po(0.01 \times 250)$. ie $X \sim Po(2.50)$

It is flawless if X=0. You can find the probability using the formula for a poisson pmf:

$P(X=0) = \frac{e^{-2.5} \cdot 2.5^0}{0!} = \frac{e^{-2.5} \cdot 1}{1} = e^{-2.5} \approx 0.0821$

This is the probability TKHunny gave you in post #2.

Step 2 Answer the question given
"You take 20 independent samples. Each is flawless with probability $e^{-2.5} \approx 0.0821$. Find the probability that at least 2 are flawless."

The number of flawed samples follows a binomial distribution. $Y \sim Bi(20,e^{-0.25})$

Evaluate $P(Y \geq 2)$ using the method you have been taught.