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Sagot :
Answer:
0.319 = 31.9% probability that the wait time will be more than an additional 16 minutes
Step-by-step explanation:
To solve this question, we need to 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.
In this question:
Event A: It has already taken 11 minutes.
Event B: It will take 16 more minutes.
Exponentially distributed with an average wait time of 14 minutes.
This means that [tex]m = 14, \mu = \frac{1}{14}[/tex]
Probability of the waiting time being of at least 11 minutes:
[tex]P(A) = P(X > 11) = e^{-\frac{11}{14}} = 0.4558[/tex]
Probability of the waiting time being of at least 11 minutes, and more than an additional 16 minutes:
More than 11 + 16 = 27 minutes. So
[tex]P(A \cap B) = P(X > 27) = e^{-\frac{27}{14}} = 0.1454[/tex]
What is the probability that the wait time will be more than an additional 16 minutes?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1454}{0.4558} = 0.319[/tex]
0.319 = 31.9% probability that the wait time will be more than an additional 16 minutes
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