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ira [324]
3 years ago
6

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

more than 4 errors in the carpet
Engineering
1 answer:
Viktor [21]3 years ago
5 0

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that \mu = 0.8

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

P(X \leq 4) + P(X > 4) = 1

We want P(X > 4). Then

P(X > 4) = 1 - P(X \leq 4)

In which

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595

P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438

P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383

P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014

0.14% probability of observing more than 4 errors in the carpet

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A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

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