The time you wait in a bank until you are served by a teller is exponentially distributed with mean 10 minutes. Calculate the probability that you will have to wait between 5 and 15 minutes.
X is exponentially distributed, so the probability density function f(x) = m.exp(-m.x), where E(X) = 1/m. You're told that the expected value is 10 mins, so 1/m = 10, so m = 1/10 = 0.1. So what you need to calculate is the integral of the pdf between 5 mins and 15 mins, that is, the integral of 0.1 times exp(-x/10) between those limits. That is -exp(-1.5) + exp(-0.5), which is approximately 0.383. So the probability of your having to wait between 5 and 15 minutes is 0.383. (I think I've got that right, but do check my reasoning.)
To calculate the probability that you will have to wait between 5 and 15 minutes, we will use the exponential distribution with a mean of 10 minutes.
The probability density function (PDF) of the exponential distribution is given by:
f(x) = (1/μ) * e^(-x/μ)
where μ is the mean of the distribution.
To find the probability that the waiting time falls between 5 and 15 minutes, we need to calculate the integral of the PDF within that range.
P(5 ≤ X ≤ 15) = ∫[5, 15] f(x) dx
Substituting the values into the formula, we have:
P(5 ≤ X ≤ 15) = ∫[5, 15] (1/10) * e^(-x/10) dx
Using calculus, we can integrate this function:
P(5 ≤ X ≤ 15) = [-e^(-x/10)] evaluated from 5 to 15
P(5 ≤ X ≤ 15) = [-e^(-15/10) - (-e^(-5/10))]
Simplifying further:
P(5 ≤ X ≤ 15) = -e^(-3/2) - (-e^(-1/2))
P(5 ≤ X ≤ 15) = e^(-1/2) - e^(-3/2)
Evaluating the exponential terms:
P(5 ≤ X ≤ 15) ≈ 0.7165 - 0.2231
P(5 ≤ X ≤ 15) ≈ 0.4934
Therefore, the probability that you will have to wait between 5 and 15 minutes is approximately 0.4934.