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kiruha [24]
3 years ago
5

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

what is the probability (a) that there will be exactly 15 arrivals? (b) no more than 10 arrivals?
Mathematics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

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.

If the mean number of arrivals is 10

This means that \mu = 10

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

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

P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is P(X \leq 10)

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

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

P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045

P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045

P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023

P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076

P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189

P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378

P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631

P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901

P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126

P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251

P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831

58.31% probability that there are no more than 10 arrivals.

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