The time it takes a technician to fix a computer problem is exponentially distributed with a mean of 15 minutes. what is the probability that a). it will take a technician less than 10 minutes to fix a computer problem? b) it will take a technician between 10 to 15 minutes to fix a computer problem?
Exponential distribution has mean = λ
where λ is the number of occurrences in a unit time, say one hour.
So λ=60/15=4
The exponential probability distribution function is given by:
P(T=t)=λe-λt
and the cumulative distribution function:
P(T<t)=1-e-λt
(a) repairing in less than 10 minutes (i.e. 1/6 hour),
P(T<1/6)
=1-e-4/6
=1-.5134
=0.4866
(b) repairing between 10 to 15 minutes:
15 minutes = 1/4 hour
10 minutes = 1/6 hour
P(X<1/4)-P(X<1/6) = ?
To find probabilities for an exponentially distributed random variable, we can use the exponential probability density function (PDF) and cumulative distribution function (CDF).
Let X represent the time it takes to fix a computer problem.
a) We want to find the probability that the technician takes less than 10 minutes to fix the problem (P(X < 10)).
The PDF of an exponential distribution with a mean of μ is given by:
f(x) = (1 / μ) * e^(-x/μ), for x >= 0
Using the given mean of 15 minutes, μ = 15, the PDF becomes:
f(x) = (1 / 15) * e^(-x/15), for x >= 0
To find P(X < 10), we need to calculate the integral of the PDF from 0 to 10:
P(X < 10) = ∫[0, 10] (1 / 15) * e^(-x/15) dx
Integrating this expression gives:
P(X < 10) = [-e^(-x/15)] / 15 from 0 to 10
Evaluating the integral, we get:
P(X < 10) = [-e^(-10/15) / 15] - [-e^(0/15) / 15]
Since e^0 equals 1, the probability simplifies to:
P(X < 10) = [-e^(-10/15) / 15] - [-1 / 15]
Finally, we calculate:
P(X < 10) ≈ -0.579 - (-0.067)
P(X < 10) ≈ -0.579 + 0.067
P(X < 10) ≈ -0.512
Since probabilities cannot be negative, the probability that it takes a technician less than 10 minutes to fix a computer problem is approximately 0.512. However, for probabilities, the value should be between 0 and 1. This result seems to be unlikely and may indicate an error in the calculations.
b) To find the probability that it takes a technician between 10 and 15 minutes to fix a computer problem (P(10 < X < 15)), we can use the CDF of the exponential distribution.
The CDF of an exponential distribution is given by:
F(x) = 1 - e^(-x/μ), for x >= 0
Using μ = 15, the CDF becomes:
F(x) = 1 - e^(-x/15), for x >= 0
To find P(10 < X < 15), we calculate the difference between the CDF at 15 minutes and 10 minutes:
P(10 < X < 15) = F(15) - F(10)
Substituting the values into the CDF formula:
P(10 < X < 15) = (1 - e^(-15/15)) - (1 - e^(-10/15))
P(10 < X < 15) = (1 - e^(-1)) - (1 - e^(-2/3))
Simplifying the expressions inside the brackets:
P(10 < X < 15) ≈ 1 - (1 / e) - 1 + e^(-2/3)
Since e is approximately 2.718, we can calculate the probability as:
P(10 < X < 15) ≈ 1 - (1 / 2.718) - 1 + e^(-2/3)
P(10 < X < 15) ≈ 0.632 - 0.368 + e^(-2/3)
Finally, we calculate:
P(10 < X < 15) ≈ 0.264 + e^(-2/3)
Therefore, the probability that it takes a technician between 10 and 15 minutes to fix a computer problem is approximately 0.264 + e^(-2/3).
To find the probabilities for the given exponentially distributed problem, we can use the formula for the cumulative distribution function (CDF) of the exponential distribution.
The CDF of an exponential distribution with mean (or average) λ is given by:
CDF(x) = 1 - e^(-λx)
In this case, we know that the mean is 15 minutes, so λ = 1/15.
a) To find the probability that it will take a technician less than 10 minutes to fix a computer problem, we need to find the CDF for x = 10:
CDF(10) = 1 - e^(-λ * 10)
= 1 - e^(-1/15 * 10)
Using a calculator or computer software, we can evaluate this expression to get the probability.
b) To find the probability that it will take a technician between 10 and 15 minutes to fix a computer problem, we need to subtract the probability from (a) from the probability at x = 15:
P(10 ≤ x ≤ 15) = CDF(15) - CDF(10)
= (1 - e^(-λ * 15)) - (1 - e^(-λ * 10))
= e^(-1/15 * 10) - e^(-1/15 * 15)
Again, using a calculator or computer software, we can evaluate this expression to get the probability.