Question

A statistics professor receives an average of five e-mail messages per day from students. assume the number of messages approximates a poisson distribution. what is the probability that on a randomly selected day she will have no messages?

Answer 1

The probability that on a randomly selected day she will have no messages $\boxed{0.00674}.$

Further Explanation:

The random variable X follows Poisson distribution.

$\boxed{X\~{\text{Poisson}}\left( \lambda \right)}$

Here, $\lambda$ represents the Poisson parameter.

The mean of the Poisson parameter is $\lambda.$

The variance of the Poisson parameter is $\lambda.$

The formula for the probability of Poisson distribution can be expressed as follows,

$\boxed{P\left( x \right)=\frac{{{e^{- \lambda }}\times {\lambda ^x}}}{{x!}}}$

Given:

The number of messages follows Poisson distribution.

The value of $\lambda$ is $\lambda = 5.$

Explanation:

A statistics professor receives an average of five e-mail messages per day from students.

The mean of the message is 5.

The probability that on a randomly selected day she will have no messages can be obtained as follows,

$\begin{aligned}P\left( 0 \right) &= \frac{{{e^{ - 5}} \times {5^0}}}{{0!}}\\&= \frac{{0.00674 \times 1}}{1}\\&= 0.00674\\\end{aligned}$

The probability that on a randomly selected day she will have no messages $\boxed{0.00674}.$

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949

2. Learn more about standard normal distribution brainly.com/question/13006989

3. Learn more about confidence interval of meanhttps://brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter: Poissondistribution

Keywords: statistics professor, receives, average, five e-mail, message, students, messages per day, Poisson distribution, probability, randomly selected, no message, parameter, probability formula, number of message.

Answer 2

From the problem, we have the following given
u = 5
x = 0

We use the Poisson distribution probability formula:
P = (e^-μ) (μ^x) / x! 
Substituting
P = (e^-5) (5^0) / 0!
P = 0.0067

The probability is 0.0067 or 0.67%