Answer:
Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is [tex]X \sim Poisson(\lambda=24.7)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
The best answer for this case would be:
C. Poisson distribution
Step-by-step explanation:
Let X the random variable that represent the number of emails from students the day before the midterm exam. For this case the best distribution for the random variable X is [tex]X \sim Poisson(\lambda=24.7)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
And for this case we want to calculate this probability:
[tex]P(X \geq 10) [/tex]
The best answer for this case would be:
C. Poisson distribution
Solution :
Given that ,
mean = [tex]\mu[/tex] = 24.7
P(X [tex]\geq[/tex]10)
C) This is the poisson distribution .
