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Sagot :
Answer:
0.207 = 20.7% probability that the wait time will be more than an additional 41 minutes
Step-by-step explanation:
To solve this question, we should understand the exponential distribution and conditional probability.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] is the probability of both A and B happening.
P(A) is the probability of A happening.
The average waiting time for this distribution is 38 minutes.
This means that [tex]m = 38, \mu = \frac{1}{38} = 0.0263[/tex]
Given that it has already taken 37 minutes, what is the probability that the wait time will be more than an additional 41 minutes?
Event A: Taking more than 37 minutes.
Event B: More than 37 + 41 = 78 minutes.
Probability of taking more than 37 minutes:
[tex]P(A) = P(X \leq 37) = 1 - e^{-0.0263*37} = 0.6221[/tex]
More than 37 minutes and more than 78 minutes:
The intersection is more than 78 minutes, so:
[tex]P(A \cap B) = P(X > 78) = e^{-0.0263*78} = 0.1286[/tex]
The probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1286}{0.6221} = 0.207[/tex]
0.207 = 20.7% probability that the wait time will be more than an additional 41 minutes
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