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In a hospital, births occur randomly at an average rate of 2 births per hour. Calculate the probability of observing

1.     3 births in a given hour at the hospital.

2.     No birth in a given hour at the hospital."

 

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Question 1

Find the centered average and trends in the following data:

Quarter

Actual

Moving Average

Centered Average

Trends

1

143

2

56

3

164

141

4

208

138

1

132

137

2

52

140

3

176

138

4

198

137

1

126

135

2

44

132

3

164

129

4

188

Let X = No birth in a given hour at the hospital

Events occur randomly

Mean rate λ = 2 ⇒ X ∼ Po (2) 

We want P(X ≥ 2) = P(X = 2) + P(X = 3) + ...

i.e. an infinite number of probabilities to calculate

 but

P(X ≥ 2) = P(X = 2) + P(X = 3) + ...

 = 1 − P(X < 2)

 = 1 − (P(X = 0) + P(X = 1))

= 1 − (e−1.8 1.80 0! + e−1.8 1.81 1! )

 = 1 − (0.16529 + 0.29753)

 = 0.537

yeh theak ans kia

2.1 Examples
Births in a hospital occur randomly at an average rate of 1.8 births per hour.
What is the probability of observing 4 births in a given hour at the hospital?
Let X = No. of births in a given hour
(i) Events occur randomly
(ii) Mean rate λ = 1.8
⇒ X ∼ Po(1.8)
We can now use the formula to calculate the probability of observing exactly 4
births in a given hour
P(X = 4) = e−1.8 1.84
4!
= 0.0723

Please let me know the handout lecture or page no. just for the reference??

Answer

100% correct

1-   0.180447

2-   0.135335

Dear, i have different answers how you're saying its 100 % correct?

Please share your comments OR solution if possible

P(X=x)=

at page 279

Solution do bhay Sohail please

please tell me handout lecture or page no. just for the reference??

 

P(X=x)=

at page 279

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