No. f is a letter of the Roman alphabet. It cannot be a probability density function.
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The formula, if any, depends on the probability distribution function for the variable. In the case of a discrete variable, X, this defines the probability that X = x. For a continuous variable, the probability density function is a continuous function, f(x), such that Pr(a < X < b) is the area under the function f, between a and b (or the definite integral or f, with respect to x, between a and b.
The exponential distribution is a continuous probability distribution with probability density definded by: f(x) = ke-kx for x ≥ 0 and f(x) = 0 otherwise.
That's a Gaussian distribution.
The answer depends on what you are trying to predict. Suppose you have a discrete random variable X with a probability density function p(X) = prob(X = x), then the expected value of a function f(X) of X is the sum of f(x)*p(x), summed over all possible values of x. For a continuous variable, the procedure is similar, except that you need to integrate rather than sum.
It the the probability that the random variable in question takes any value up to and including the argument. Suppose you have a random variable X and f(x) is the probability that X = x [that is, the rv X takes the value x]. If F(x) denotes the cumulative distribution function of X, then F(x) is the sum of all f(y) where y <= x. Thus, for a fair die, F(1) = f(1) = 1/6 F(2) = f(1) + f(2) = 2/6 F(3) = f(1) + f(2) + f(3) = 3/6 and so on. Note that F(X) = 0 for X < 1, F(a+b) where a is an integer in the interval [1,6] and 0<b<1 is F(a). Thus, for example, F(3.5) = F(3). and F(x) = 1 for x >=6. In the case of continuous probability distributions, the summation is replaced by integration.