Question

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour. What is the probability that in one hour more than 5 clients arrive

Answer 1

Answer:

1.76% probability that in one hour more than 5 clients arrive

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.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

Answer 2

Answer:

Probability that in one hour more than 5 clients arrive is 0.0166.

Step-by-step explanation:

We are given that the arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

The Poisson distribution is foe random variable is given by;

     $P(X=x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; x = 0,1,2,3,....$

where, $\lambda$ = rate of arrival per hour =2

Let X = Arrival of clients

So, Probability that in one hour more than 5 clients arrive = P(X > 5)

P(X > 5) = $1-P(X \leq 5)$

P(X $\leq$ 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)  

              = $\frac{e^{-2} \times 2^{0} }{0!} +\frac{e^{-2} \times 2^{1} }{1!} +\frac{e^{-2} \times 2^{2} }{2!} +\frac{e^{-2} \times 2^{3} }{3!} +\frac{e^{-2} \times 2^{4} }{4!} +\frac{e^{-2} \times 2^{5} }{5!}$

              = 0.1353 + 0.2707 + 0.2707 + 0.1804 + 0.0902 + 0.0361

              = 0.9834

So, P(X > 5) = $1-P(X \leq 5)$ = 1 - 0.9834 = 0.0166

Therefore, probability that in one hour more than 5 clients arrive is 0.0166.