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(Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.3.
Q. The probability that there will be five or more murder in a given week is
- a)0.1847
- b)0.2461
- c)0.3927
- d)0.4167
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
(Q.3-Q.5) Delhi averages three murder per week and their occurrences f...
P(5 or more) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 0.1847.
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(Q.3-Q.5) Delhi averages three murder per week and their occurrences f...
To find the probability that there will be five or more murders in a given week in Delhi, we can use the Poisson distribution. The Poisson distribution is commonly used to model the number of events occurring in a fixed interval of time or space.
The Poisson distribution is characterized by a single parameter, λ (lambda), which represents the average number of events in the given interval. In this case, we are given that Delhi averages three murders per week, so λ = 3.
To calculate the probability of five or more murders, we need to calculate the cumulative probability of zero to four murders and subtract it from 1.
Let's calculate the probability step by step:
1. Calculate the probability of zero to four murders using the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where X is the random variable representing the number of murders and k is the number of murders we want to calculate the probability for.
For k = 0:
P(X = 0) = (e^(-3) * 3^0) / 0! = e^(-3) ≈ 0.0498
For k = 1:
P(X = 1) = (e^(-3) * 3^1) / 1! = 3e^(-3) ≈ 0.1493
For k = 2:
P(X = 2) = (e^(-3) * 3^2) / 2! = 9e^(-3) / 2 ≈ 0.2240
For k = 3:
P(X = 3) = (e^(-3) * 3^3) / 3! = 27e^(-3) / 6 ≈ 0.2240
For k = 4:
P(X = 4) = (e^(-3) * 3^4) / 4! = 81e^(-3) / 24 ≈ 0.1680
2. Calculate the cumulative probability of zero to four murders:
P(0 to 4 murders) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ≈ 0.0498 + 0.1493 + 0.2240 + 0.2240 + 0.1680 ≈ 0.8151
3. Subtract the cumulative probability from 1 to get the probability of five or more murders:
P(≥ 5 murders) = 1 - P(0 to 4 murders) = 1 - 0.8151 ≈ 0.1849
Therefore, the probability that there will be five or more murders in a given week in Delhi is approximately 0.1849, which corresponds to option A.
The Poisson distribution is characterized by a single parameter, λ (lambda), which represents the average number of events in the given interval. In this case, we are given that Delhi averages three murders per week, so λ = 3.
To calculate the probability of five or more murders, we need to calculate the cumulative probability of zero to four murders and subtract it from 1.
Let's calculate the probability step by step:
1. Calculate the probability of zero to four murders using the Poisson distribution formula:
P(X = k) = (e^(-λ) * λ^k) / k!
Where X is the random variable representing the number of murders and k is the number of murders we want to calculate the probability for.
For k = 0:
P(X = 0) = (e^(-3) * 3^0) / 0! = e^(-3) ≈ 0.0498
For k = 1:
P(X = 1) = (e^(-3) * 3^1) / 1! = 3e^(-3) ≈ 0.1493
For k = 2:
P(X = 2) = (e^(-3) * 3^2) / 2! = 9e^(-3) / 2 ≈ 0.2240
For k = 3:
P(X = 3) = (e^(-3) * 3^3) / 3! = 27e^(-3) / 6 ≈ 0.2240
For k = 4:
P(X = 4) = (e^(-3) * 3^4) / 4! = 81e^(-3) / 24 ≈ 0.1680
2. Calculate the cumulative probability of zero to four murders:
P(0 to 4 murders) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) ≈ 0.0498 + 0.1493 + 0.2240 + 0.2240 + 0.1680 ≈ 0.8151
3. Subtract the cumulative probability from 1 to get the probability of five or more murders:
P(≥ 5 murders) = 1 - P(0 to 4 murders) = 1 - 0.8151 ≈ 0.1849
Therefore, the probability that there will be five or more murders in a given week in Delhi is approximately 0.1849, which corresponds to option A.
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(Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.3.Q. The probability that there will be five or more murder in a given week isa)0.1847b)0.2461c)0.3927d)0.4167Correct answer is option 'A'. Can you explain this answer?
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(Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.3.Q. The probability that there will be five or more murder in a given week isa)0.1847b)0.2461c)0.3927d)0.4167Correct answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about (Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.3.Q. The probability that there will be five or more murder in a given week isa)0.1847b)0.2461c)0.3927d)0.4167Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for (Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.3.Q. The probability that there will be five or more murder in a given week isa)0.1847b)0.2461c)0.3927d)0.4167Correct answer is option 'A'. Can you explain this answer?.
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