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castortr0y [4]
3 years ago
15

3/4(8p+12)+3/8(8-16p)

Mathematics
1 answer:
nikitadnepr [17]3 years ago
4 0
3/4(8p+12)+3/8(8-16p)
6p+9+3-6p
6p-6p=0
9+3=12 
so p=12
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A certain kind of sheet metal has, on average, 3 defects per 18 square feet. Assuming a Poisson distribution, find the probabili
Fofino [41]

Answer:

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

Step-by-step explanation:

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 time interval.

In this problem, we have that:

3 defects per 18 square feet.

So for 31 square feet, we have to solve a rule of three

3 defects - 18 square feet

x defects - 31 square feet

18x = 3*31

x = \frac{31*3}{18}

x = 5.17

So \mu = 5.17

Assuming a Poisson distribution, find the probability that a 31 square foot metal sheet has at least 4 defects.

Either it has three or less defects, or it has at least 4 defects. The sum of the probabilities of these events is decimal 1.

So

P(X \leq 3) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X \leq 3)

In which

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

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

P(X = 0) = \frac{e^{-5.17}*(5.17)^{0}}{(0)!} = 0.0057

P(X = 1) = \frac{e^{-5.17}*(5.17)^{1}}{(1)!} = 0.0294

P(X = 2) = \frac{e^{-5.17}*(5.17)^{2}}{(2)!} = 0.0760

P(X = 3) = \frac{e^{-5.17}*(5.17)^{3}}{(3)!} = 0.1309

So

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0057 + 0.0294 + 0.0760 + 0.1309 = 0.242

Finally

P(X \geq 4) = 1 - P(X \leq 3) = 1 - 0.242 = 0.758

75.8% probability that a 31 square foot metal sheet has at least 4 defects.

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