Question

A company produces optical-fiber cable with a mean of 0.6 flaws per 100 feet. What is the probability that there will be exactly 4 flaws in 1000 feet of cable

Answer 1

The probability of finding exactly 4 flaws in 1000 feet of an optical-fiber cable with an average of 0.6 flaws per 100 feet is determined using the Poisson distribution formula.

Step-by-step explanation:

The question asks about the probability of finding exactly 4 flaws in 1000 feet of optical-fiber cable given a mean of 0.6 flaws per 100 feet. To find this probability, we can use the Poisson distribution formula:

P(X=k) = (e^(-λ) * λ^k) / k!

Where:

e is the base of the natural logarithm (approximately 2.71828),

λ (lambda) is the average rate of occurrence (the expected number of occurrences over the interval),

k is the number of occurrences for which we're finding the probability,

k! is the factorial of k.

In this case:

λ = 0.6 flaws per 100 feet * 10 (since we're considering 1000 feet) = 6 flaws,

k = 4 (the number of flaws we want to find the probability for).

So, the probability P(X=4) is:

P(X=4) = (e^(-6) * 6^4) / 4!

After calculating the above expression, we'll get the probability that there are exactly 4 flaws in 1000 feet of optical-fiber cable. This type of calculation is common in statistics and is crucial for quality control in manufacturing and engineering contexts.

Answer 2

`P(X = 4) =0.133`

Step-by-step explanation:

From the question we are told that:

Mean
`\x=0.6/100`

Flaws
`f=4`

Distance
`d_2=1000ft`

Generally the equation for Poisson mean lambda over 1000 is mathematically given by

`\lambda=0.6*1000/100`

`\lambda = 6`

Therefore

`P(X = 4) = {e^-\lambda * \lambda^(\=x)} {{\=x}!}`

`P(X = 4) =( e^(-6) * 6^4)/( 4!)`

`P(X = 4) =0.133`