Respuesta :
Answer:
0.11711612445;
0.99376780489;
2.4494897
Step-by-step explanation:
Given that :
Arrival rate = 6 per hour
Mean Arrival during 90 minute period :
60 min = 6
90 min = (6 * 90) / 60 = 9
λ = 9
Using poisson :
P(x =x) = (e^-λ * λ^x) / x!
1) Exactly 7
P( x = 7) = (e^-9 * 9^7) / 7!
= 590.26526724266 / 5040
= 0.11711612445
Atleast 3:
P(x ≥ 3)
P(x ≥ 3) = 1 - P(x < 3)
P(x < 3) = p(x =0) + p(x = 1) + p(x = 2)
P(x = 0) = (e^-9 * 9^0) / 0! = 0.0001234098
P(x = 1) = (e^-9 * 9^1) / 1! = 0.00111068824
P(x = 2) = e^-9 * 9^2) / 2! = 0.00499809707
1 - P(x < 3) =
1 - (0.0001234098 + 0.00111068824 + 0.00499809707)
1 - 0.00623219511
= 0.99376780489
B.) mean and standard deviation of the number of patients arriving at the hospital accident and emergency department in an hour?
Mean = arrival per hour = λ = 6
Standard deviation = √λ = √6
Standard deviation = 2.4494897
The probability that, during any 90-minute period, the number of patients  arriving at the hospital accident and emergency department is exactly 7 is 0.11711612445 and the probability for at least 3 is 0.99376780489. The mean is 6 and the standard deviation is 2.4494897.
Given :
Patients arrive at a hospital accident and emergency department at random follow a  Poisson distribution at a rate of 6 per hour.
Accprding to the poisson distribution:
[tex]\rm P(x=x)=\dfrac{(e^{-\lambda} \times \lambda^x )}{x!}[/tex]
a) The probability that, during any 90-minute period, the number of patients  arriving at the hospital accident and emergency department is:
First, determine the value of [tex]\lambda[/tex]:
[tex]\lambda = \dfrac{6\times 90}{60} = 9[/tex]
(i) At (x = 7) the probability is given by:
[tex]\rm P(x=7)=\dfrac{(e^{-9} \times 9^7 )}{7!}[/tex]
Simplify the above expression.
P(x = 7) = 0.11711612445
(ii) At (x [tex]\geq[/tex] 3) the probability is given by:
[tex]\rm P(x\geq 3)=1-P(x<3)[/tex]
[tex]\rm P(x\geq 3)=1-P(x=2)-P(x=1)-P(x=0)[/tex]
[tex]\rm P(x=2)=\dfrac{(e^{-9} \times 9^2 )}{2!}=0.00499809707[/tex]
[tex]\rm P(x=1)=\dfrac{(e^{-9} \times 9^1 )}{1!}= 0.00111068824[/tex]
[tex]\rm P(x=0)=\dfrac{(e^{-9} \times 9^0 )}{0!}= 0.0001234098[/tex]
So, the probability at (x [tex]\geq[/tex] 3) is given by:
[tex]\rm P(x\geq 3)=1-0.0001234098 - 0.00111068824 - 0.00499809707[/tex]
[tex]\rm P(x\geq 3)=1 - 0.00623219511[/tex]
[tex]\rm P(x\geq 3)= 0.99376780489[/tex]
b) The mean and standard deviation of the number of patients arriving at the  hospital accident and emergency department in an hour are given by:
[tex]\rm Mean = \lambda = 6\\\\[/tex]
[tex]\rm Standard\; Deviation=\sqrt{\lambda }= \sqrt{6} = 2.4494897[/tex]
For more information, refer to the link given below:
https://brainly.com/question/17716064
