Answer:
a) It can be used because np and n(1-p) are both greater than 5.
Step-by-step explanation:
Binomial distribution and approximation to the normal:
The binomial distribution has two parameters:
n, which is the number of trials.
p, which is the probability of a success on a single trial.
If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.
In this question:
So, lets verify the conditions:
np = 201*0.45 = 90.45 > 5
n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5
Since both np and n(1-p) are greater than 5, the approximation can be used.
Answer:
4
Step-by-step explanation:
I believe the answer is B because you have to do the Vertical Line Test.<span />
Answer:
The steady state proportion for the U (uninvolved) fraction is 0.4.
Step-by-step explanation:
This can be modeled as a Markov chain, with two states:
U: uninvolved
M: matched
The transitions probability matrix is:
The steady state is that satisfies this product of matrixs:
![[\pi] \cdot [P]=[\pi]](https://tex.z-dn.net/?f=%5B%5Cpi%5D%20%5Ccdot%20%5BP%5D%3D%5B%5Cpi%5D)
being π the matrix of steady-state proportions and P the transition matrix.
If we multiply, we have:
Now we have to solve this equations
We choose one of the equations and solve:
Then, the steady state proportion for the U (uninvolved) fraction is 0.4.
