Where is the additional lambda while solving problems for an Exponential distribution?

Question

I've been studying the problems for an exponential distribution and I have this somewhat basic question.

Known

The pdf in an exponential distribution where $x \ge 0$ is given by:

$$ f(x) = \lambda e^{-\lambda x} $$

Example Problem

Suppose that the amount of time one spends in a bank is exponentially distributed with mean ten minutes, that is, λ = 1/10 . What is the probability that a customer will spend more than fifteen minutes in the bank?

Example Solution

If X represents the amount of time that the customer spends in the bank, then the probability is just:

$$P\{X > 15\} = e^{-15\lambda} = e^{-3/2}$$

My question

Where is the other $\lambda$ from the pdf ? (i.e. why did we take the solution above to be $e^{-15\lambda}$ and not $\lambda e^{-15\lambda}$ ?)

Answer 1

$\lambda e^{-\lambda x}$ is the density function. The distribution function is given by $P(X \leq x) =1-e^{-\lambda x}$ which gives $P(X>x)=e^{-\lambda x}$. $\lambda e^{-\lambda x}$ is the derivative of $1-e^{-\lambda x}$.

Answer 2

We need the area under of all of the curve to the right of $15$. The indefinite integral is $-e^{-\lambda x}$. The definite integral is $F(\infty) - F(x)$. Since $-e^{-\infty}=0$, this simplifies to gives us $e^{-\lambda x}$.

In fact, the reason we have the extra $\lambda$ in the PDF is to get it to integrate to $1$. Without the $\lambda$, the integral of the PDF from $0$ to $\infty$ would be $\frac 1 {\lambda}$. Since the total probability has to be $1$, we have to normalize the PDF.