Respuesta :
Using the Poisson distribution, it is found that there is a 0.1755 = 17.55% probability that she received exactly 4 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday.
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
- [tex]\mu[/tex] is the mean in the given interval.
In this problem, there is an average of 5 non-work emails per hour, for any one-hour interval, thus [tex]\mu = 5[/tex].
The probability of receiving exactly 4 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday is P(X = 4), since 4 pm to 5 pm is a one-hour interval. Hence:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 4) = \frac{e^{-5}5^{4}}{(4)!} = 0.1755[/tex]
0.1755 = 17.55% probability that she received exactly 4 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday.
A similar problem is given at https://brainly.com/question/24230648
