Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter λ (lambda) = 0.5.
What's the probability that a repair takes less than 5 hours? AND what's the conditional probability that a repair takes at least 11 hours, given that it takes more than 8 hours?
A factory produces 64 Xboxes in 3 hours. Find how many hours it takes to produce 16 Xboxes at this rate?
To find the probability that a repair takes less than 5 hours, we can use the cumulative distribution function (CDF) of the exponential distribution. The CDF of an exponential distribution with parameter λ is given by:
F(x) = 1 - e^(-λx)
where x is the value at which we want to evaluate the CDF.
1. Probability that a repair takes less than 5 hours:
Let's substitute λ = 0.5 and x = 5 into the CDF formula:
F(5) = 1 - e^(-0.5*5)
= 1 - e^(-2.5)
≈ 1 - 0.082
≈ 0.918
Therefore, the probability that a repair takes less than 5 hours is approximately 0.918, or 91.8%.
2. Conditional probability that a repair takes at least 11 hours, given that it takes more than 8 hours:
To find this conditional probability, we need to calculate the probability that a repair takes at least 11 hours and also takes more than 8 hours, and then divide it by the probability that it takes more than 8 hours.
P(X ≥ 11 | X > 8) = P(X ≥ 11 and X > 8) / P(X > 8)
Let's calculate each term of the equation separately:
P(X ≥ 11 and X > 8) = P(X ≥ 11) = 1 - F(11)
= 1 - (1 - e^(-0.5*11))
= e^(-0.5*11)
≈ 0.082
P(X > 8) = 1 - F(8)
= 1 - (1 - e^(-0.5*8))
= e^(-0.5*8)
≈ 0.393
Now let's divide the probability that the repair takes at least 11 hours by the probability it takes more than 8 hours:
P(X ≥ 11 | X > 8) ≈ (e^(-0.5*11)) / (e^(-0.5*8))
≈ 0.082 / 0.393
≈ 0.209
Therefore, the conditional probability that a repair takes at least 11 hours, given that it takes more than 8 hours, is approximately 0.209, or 20.9%.
To find the probability that a repair takes less than 5 hours, we need to calculate the cumulative distribution function (CDF) of the exponential distribution with λ = 0.5.
The CDF of an exponential distribution is given by the formula:
CDF(x) = 1 - e^(-λx)
Substituting λ = 0.5 and x = 5 into the CDF formula:
CDF(5) = 1 - e^(-0.5 * 5)
= 1 - e^(-2.5)
≈ 0.9179
Therefore, the probability that a repair takes less than 5 hours is approximately 0.9179, or 91.79%.
Now, to find the conditional probability that a repair takes at least 11 hours, given that it takes more than 8 hours, we need to use the conditional probability formula:
P(A | B) = P(A ∩ B) / P(B)
Where A is the event "repair takes at least 11 hours" and B is the event "repair takes more than 8 hours".
First, let's find P(B), the probability that a repair takes more than 8 hours. We can calculate it by subtracting the CDF(8) from 1:
P(B) = 1 - CDF(8)
= 1 - (1 - e^(-0.5 * 8))
= e^(-4)
≈ 0.0183
Next, we need to find P(A ∩ B), the probability that a repair takes at least 11 hours and more than 8 hours. This can be calculated by subtracting the CDF(11) from the CDF(8):
P(A ∩ B) = CDF(8) - CDF(11)
= 1 - e^(-0.5 * 8) - (1 - e^(-0.5 * 11))
≈ 0.0172
Finally, we can calculate the conditional probability:
P(A | B) = P(A ∩ B) / P(B)
= 0.0172 / 0.0183
≈ 0.9406
Therefore, the conditional probability that a repair takes at least 11 hours, given that it takes more than 8 hours, is approximately 0.9406, or 94.06%.
The probability distribution you are talking about is
f (x; lambda) = lambda exp^-(lambda x), x > 0
It is discussed at http://en.wikipedia.org/wiki/Exponential_distribution
The probability that a repair takes less than x is
F (x; lambda) = 1 - exp(-lambda x)
For x = 1 - exp(-2.5) = 0.918
The fraction taking 8 hours or more to repair is
exp(-4) = 0.01832
The fraction taking 11 hours to repair is
exp(-5.5) = 0.00409
That means that 409/1832 = 22% of the repairs taking 8 hours or more require more than 11 hours.