Question

Messages arrive at an electronic message center at random times, with an average of 9 messages per hour. (a) What is the probability of receiving at least five messages during the next hour? (b) What is the probability of receiving exactly five messages during the next hour?

Answer 1

Answer:

The probability of receiving at least five messages during the next hour is 0.945

The probability of receiving exactly five messages during the next hour is 0.0607

Step-by-step explanation:

We are given that Messages arrive at an electronic message center at random times, with an average of 9 messages per hour.

X follows poisson distribution with $\lambda = 9$

Formula : $P(X=x)=\frac{e^{-\lambda} \lambda^x}{x!}$

a) We are supposed to find  the probability of receiving at least five messages during the next hour i.e. $P(X\geq 5)=1-P(X<5)$

1-P(X<5)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

$1-P(X<5)=1-(\frac{e^{-9} 9^0}{0!}+\frac{e^{-9} 9^1}{1!}+\frac{e^{-9} 9^2}{2!}+\frac{e^{-9} 9^3}{3!}+\frac{e^{-9} 9^4}{4!})$

1-P(X<5)=0.945

So,$P(X\geq 5)=0.945$

So, The probability of receiving at least five messages during the next hour is 0.945

(b) What is the probability of receiving exactly five messages during the next hour?

$P(X=5)=\frac{e^{-9} 9^5}{5!}$

P(X=5)=0.0607

Hence the probability of receiving exactly five messages during the next hour is 0.0607

Answer 2

This question is based on the probability.Therefore the answer is,(a) $P(X\geq 5) = 0.945$ and (b) = $P(X=5) = 0.0607$.

Given:

Messages arrive at an electronic message center at random times, with an average of 9 messages per hour.

According to the question,

Now, X follows the poison distribution with,

$P(X=x) = \dfrac{e^{- \lambda } \lambda^x}{x!}$

a) We need to find  the probability of receiving at least five messages during the next hour i.e.

$P(X\geq 5) = 1 - P(X<5)$

1 - P(X<5) = 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

$1-P(X<5)=1-(\dfrac{e^{- 9} 9^0}{0!} + \dfrac{e^{- 9} 9^1}{1!}+\dfrac{e^{- 9} 9^2}{2!}+\dfrac{e^{- 9} 9^3}{3!}+\dfrac{e^{- 9} 9^4}{4!})$

1 - P(X<5)=0.945

So, $P(X\geq 5) = 0.945$

Thus, The probability of receiving at least five messages during the next hour is 0.945.

 

(b) Now, find he probability of receiving exactly five messages during the next hour is,

   $P(X=5) = \dfrac{e^{- 9} 9^5}{5!}\\P(X=5) = 0.0607$

Thus, the probability of receiving exactly five messages during the next hour is 0.0607.

Therefore the answer is,(a) $P(X\geq 5) = 0.945$ and (b) = $P(X=5) = 0.0607$.

For more details, prefer this link:

https://brainly.com/question/11234923