Question #3929

A rail company knows that on average they will have 3 people ill each day. In order to make plans for replacements they have asked you to calculate the following probabilities: In a single day - (1) No employees ill (2)Two or less employees ill (3) Four or more employees ill

Expert's answer

The unknown probabilities depend on distribution of number of number of ill patients.

But in such situations the most frequently is used Poisson distribution.

1) No employees ill, X = 0, lambda=3

P(0) = e^{-3} 3^{0}/0! = 0.5*1/1 = .05

2)Two or less ill

P(X<=2) = P(0) + P(1) + P(2) = .15

P(1) = e^{-3} 3^{1}/1! = 0.05*3/1 = 0.15

P(2) = e^{-3} 3^{2}/2! = 0.05*9/2 = 0.22

P(X<=2) = P(0) + P(1) + P(2)= .05 +.15 + .22 =.42

3)Probability of four or more employees ill

P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3))

P(3) = e^{-3} 3^{3}/3! = 0.05*27/6 = 0.27

P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3)) = 1 – (.05 +.15 +.22 +.27) = 1 - .69 =.31

But in such situations the most frequently is used Poisson distribution.

1) No employees ill, X = 0, lambda=3

P(0) = e

2)Two or less ill

P(X<=2) = P(0) + P(1) + P(2) = .15

P(1) = e

P(2) = e

P(X<=2) = P(0) + P(1) + P(2)= .05 +.15 + .22 =.42

3)Probability of four or more employees ill

P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3))

P(3) = e

P(X>=4) = 1 – P(X<=3) = 1 – (P(0) + P(1) + P(2) + P(3)) = 1 – (.05 +.15 +.22 +.27) = 1 - .69 =.31

## Comments

Assignment Expert16.03.12, 17:55Dear Craig

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Craig11.08.11, 21:05Thank you so much

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