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Question 1 :

The time to service a customer at a bank teller's counter is exponentially distributed with mean of 60 seconds. What is the probability that the three customers in the front of an arriving customer that each will take less than 100 seconds to complete their transactions?

Question 2 :

For exponentially distributed random variable X, find the value of $\lambda$ that satisfies the following equation

$$P( X \le 3 ) = 0.9 \cdot P( X \le4 )$$

For the 1st question I can find value of $\lambda$, i.e. Mean $E(X)=1/\lambda$. For the probability should I use pdf or cdf of exponential distribution ?

For the 2nd question, how to separate $\lambda$, i.e. bring it to RHS given

$$1-e^{-3\lambda} = 0.9 \cdot \bigl(1 - e^{-4*\lambda}\bigr)$$

Please help with these two questions.

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Is it the sum of the service times that is less than $100$, or do we want each of the three service times to be less than $100$? The second version is quite a bit simpler than the first. –  André Nicolas Oct 8 '12 at 15:56
    
Yes each of the service times to be less than 100. So can you provide the solution for the 2nd question. –  RAZER Oct 9 '12 at 5:22
    
For first question, answer is then $(1-e^{-100/60})^3$. For second problem, multiply through by $e^{4\lambda}$, and let $y=e^\lambda$. Or leave things alone and let $z=e^{-\lambda}$. In either case we get a fourth degree equation. In either case, $y=1$ (or $z=1$) is a solution (not the intended one). Reduce, we get a cubic. In the case of $y=e^{\lambda}$ the cubic has one positive root. It is not a "nice" root, we have to use a numerical method, which we might as well apply instead to the original quartic. Can for example use Newton Method. Then take the logarithm to find $\lambda$. –  André Nicolas Oct 9 '12 at 5:35
    
Cubics equations (and quartics) do have "formulas" for the roots, but they are usually not computationally useful. Instead of Newton-Raphson, can ask Wolfram Alpha for the roots, or some other software, even the "Solve" button present on some calculators. –  André Nicolas Oct 9 '12 at 5:40
    
Thank you very much, now i got it –  RAZER Oct 9 '12 at 5:52
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2 Answers

up vote 1 down vote accepted

It's not a question of whether you use the PDF or CDF. These are not interchangeable functions. Let's look at what the PDFs and CDFs mean.

The cumulative distribution function (CDF) gives you the probability that a variable is less than a certain value. For a single random variable, $x$, as $x \to \infty$ then $\textrm{CDF}(x) \to 1$. That is, the probability of a random variable being less than some arbitrary given value tends to 100% as you make that arbitrary given value very large.

The probability density function (PDF) for a continuous random variable is a strange animal. The value $\textrm{PDF}(x)$ for a continuous random variable does not mean anything. This seems strange, but remember that for a continuous random variable, $P(x = X) = 0$. That is, the probability of a continuous random variable assuming any value $X$ is exactly zero. So what's the point of a PDF?

A PDF is defined in a very specific way: namely, it must be such that the integral of the PDF from negative infinity to some value $X$ gives you exactly the probability that $x \le X$. In other words, the PDF is the integrand of the CDF -- the CDF is just the integral of the PDF!

Thus, the CDF gives you probabilities, and through the Fundamental Theorem of Calculus, the PDF gives you the characteristics of the distribution. The "bell curve" is the PDF of the normal distribution, and the area under the curve up to a certain value is the probability!

So, to answer your first question, you need to compute the CDF. If I've made this explanation clear, it's not that you choose to use one or the other -- you use both, because both are inextricably linked!

For your second question, you also need to use both: first, you integrate the PDF from $x = -\infty$ to $x = 3$, and you get an answer in terms of $\lambda$. Then, you do the same from $x = -\infty$ to $x = 4$, and get another answer in terms of lambda. Set the first equal to 0.9 times the second, and solve for $\lambda$. You have nearly done that in your response, but you did not accurately compute the probabilities.

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Q1: The distribution of sums of exponential random variables has a Gamma / Erlang distribution: http://www.johndcook.com/distribution_chart.html#gamma_exponential http://en.wikipedia.org/wiki/Gamma_distribution

For the probability you should use the CDF

Q2: It is quite lengthy, you basically have to solve a polynomial equation after multiplying with e^(4*lambda) and substituting e^lambda -> x.

Peter

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