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Mathematics

1 answer:

Rom4ik [11]3 years ago

6 0

Answer:

40-26.50=m

Step-by-step explanation:

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Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec

o-na [289]

Answer:

(a) The probability of the event (X > 84) is 0.007.

(b) The probability of the event (X < 64) is 0.483.

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 64.

The probability mass function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2, ...$

(a)

Compute the probability of the event (X > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

$=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007$

Thus, the probability of the event (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

$=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483$