Mathematics, 22.11.2020 15:50 jacobhawkins112471

2. Patients arrive at a hospital accident and emergency department at random follow a Poisson distribution at a rate of 6 per hour.
(a) Find the probability that, during any 90-minute period, the number of patients
arriving at the hospital accident and emergency department is
(i) exactly 7
(2 marks)
(ii) at least 3
(2 marks)
(b) What is the mean and standard deviation of the number of patients arriving at the
hospital accident and emergency department in an hour?
(2 marks)

Answers: 1

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Answers

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:

$\rm P(x=x)=\dfrac{(e^{-\lambda} \times \lambda^x )}{x!}$

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 $\lambda$ :

$\lambda = \dfrac{6\times 90}{60} = 9$

(i) At (x = 7) the probability is given by:

$\rm P(x=7)=\dfrac{(e^{-9} \times 9^7 )}{7!}$

Simplify the above expression.

P(x = 7) = 0.11711612445

(ii) At (x $\geq$ 3) the probability is given by:

$\rm P(x\geq 3)=1-P(x$

$\rm P(x\geq 3)=1-P(x=2)-P(x=1)-P(x=0)$

$\rm P(x=2)=\dfrac{(e^{-9} \times 9^2 )}{2!}=0.00499809707$

$\rm P(x=1)=\dfrac{(e^{-9} \times 9^1 )}{1!}= 0.00111068824$

$\rm P(x=0)=\dfrac{(e^{-9} \times 9^0 )}{0!}= 0.0001234098$

So, the probability at (x $\geq$ 3) is given by:

$\rm P(x\geq 3)=1-0.0001234098 - 0.00111068824 - 0.00499809707$

$\rm P(x\geq 3)=1 - 0.00623219511$

$\rm P(x\geq 3)= 0.99376780489$

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:

$\rm Mean = \lambda = 6\\\\$

$\rm Standard\; Deviation=\sqrt{\lambda }= \sqrt{6} = 2.4494897$

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Answer from: Callmetay24

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

Answer from: Quest

Just divide the x axis (the top) by 2 or multiply the y axis (the bottom) by 2

Answer from: Quest

Do the inverse operation

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2. Patients arrive at a hospital accident and emergency department at random follow a Poisson distr...