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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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This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle W_{net = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

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As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ^{Y-1 = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = ( T₂ / T₁ )^{\frac{1}{Y-1}

so we substitute

⇒ V₁ / V₂ = (  973 K / 303 K  )^{\frac{1}{1.4-1}

= (  3.21122  )^{\frac{1}{0.4}

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

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we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

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Since pressure is constant, V ∝ T

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cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

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