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mario62 [17]
3 years ago
6

I need some help with this

Mathematics
2 answers:
joja [24]3 years ago
8 0
The answer is (1, -2)

because f(x) = y

so, where y is -2 , x is 1
Harman [31]3 years ago
3 0
The other answer is wrong, the right answer is f(-2) = 2
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Consider a Poisson distribution with μ = 6.
bearhunter [10]

Answer:

a) P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b) f(2) = 0.04462

c) f(1) = 0.01487

d) P(X \geq 2) = 0.93803

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 successes

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

In this question:

\mu = 6

a. Write the appropriate Poisson probability function.

Considering \mu = 6

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

b. Compute f (2).

This is P(X = 2). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

P(X \geq 2) = 1 - P(X < 2)

In which:

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248

P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

Then

P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197

P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803

So

P(X \geq 2) = 0.93803

5 0
3 years ago
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