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The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019?
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The number of calls arriving at an interal switch board of an office i...
Solution:
Given that the number of calls arriving at an internal switch board of an office is 96 per hour.
We need to find the probability that there will be:
i) not more than 3 calls on the board
ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019
To solve the above problem, we need to assume that the number of calls arriving at the switchboard follows a Poisson distribution. The Poisson distribution is given by:
P(X=k) = (e^-λ * λ^k)/k!
Where,
P(X=k) is the probability of getting k calls in a given time interval
e is the mathematical constant approximately equal to 2.71828
λ is the average rate of calls arriving at the switchboard
k is the number of calls
Let us assume that λ = 96/60 = 1.6 (since there are 60 minutes in an hour)
i) Probability of not more than 3 calls on the board
We need to find the probability of getting 0, 1, 2 or 3 calls on the board.
P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= (e^-1.6 * 1.6^0)/0! + (e^-1.6 * 1.6^1)/1! + (e^-1.6 * 1.6^2)/2! + (e^-1.6 * 1.6^3)/3!
= 0.216 + 0.346 + 0.277 + 0.147
= 0.986
Therefore, the probability of not more than 3 calls on the board is 0.986.
ii) Probability of at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019
We need to find the probability of getting at least 3 calls in a minute, given e^-1.6 = 0.2019.
P(X ≥ 3) = 1 - P(X=0) - P(X=1) - P(X=2)
= 1 - (e^-1.6 * 1.6^0)/0! - (e^-1.6 * 1.6^1)/1! - (e^-1.6 * 1.6^2)/2!
= 1 - 0.2019 - 0.3231 - 0.2596
= 0.2154
Therefore, the probability of getting at least 3 calls in a minute on the board, given e^-1.6 = 0.2019, is 0.2154.
Hence, the solution is obtained.
Given that the number of calls arriving at an internal switch board of an office is 96 per hour.
We need to find the probability that there will be:
i) not more than 3 calls on the board
ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019
To solve the above problem, we need to assume that the number of calls arriving at the switchboard follows a Poisson distribution. The Poisson distribution is given by:
P(X=k) = (e^-λ * λ^k)/k!
Where,
P(X=k) is the probability of getting k calls in a given time interval
e is the mathematical constant approximately equal to 2.71828
λ is the average rate of calls arriving at the switchboard
k is the number of calls
Let us assume that λ = 96/60 = 1.6 (since there are 60 minutes in an hour)
i) Probability of not more than 3 calls on the board
We need to find the probability of getting 0, 1, 2 or 3 calls on the board.
P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= (e^-1.6 * 1.6^0)/0! + (e^-1.6 * 1.6^1)/1! + (e^-1.6 * 1.6^2)/2! + (e^-1.6 * 1.6^3)/3!
= 0.216 + 0.346 + 0.277 + 0.147
= 0.986
Therefore, the probability of not more than 3 calls on the board is 0.986.
ii) Probability of at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019
We need to find the probability of getting at least 3 calls in a minute, given e^-1.6 = 0.2019.
P(X ≥ 3) = 1 - P(X=0) - P(X=1) - P(X=2)
= 1 - (e^-1.6 * 1.6^0)/0! - (e^-1.6 * 1.6^1)/1! - (e^-1.6 * 1.6^2)/2!
= 1 - 0.2019 - 0.3231 - 0.2596
= 0.2154
Therefore, the probability of getting at least 3 calls in a minute on the board, given e^-1.6 = 0.2019, is 0.2154.
Hence, the solution is obtained.
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The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019?
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The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019?.
The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The number of calls arriving at an interal switch board of an office is 96 per hour find the probability that there will be - i) not more than 3calls on the board ii) at least 3 calls in a minute on the board given e‐¹.⁶ = 0.2019?.
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