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No. f is a letter of the Roman alphabet. It cannot be a probability density function.

Q: Could f 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 â‰¥ 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.

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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.

Suppose the probability density function is f(x), defined over a domain D Then the mean is E(X) = x*f(x) integrated with respect to x over D. Calculate E(X2) = x2*f(x) integrated with respect to x over D. Then Variance(X) = E(X2) - [E(X)]2 and Standard Deviation = sqrt(Variance).

In terms of probability theory, the cumulative distribution function (cdf) is the result of the summation or integration of the probability density function (pdf). The cdf F(a) is the area under the pdf from its lower limit to a. I hope I am responding to your question. If not, perhaps you can clarify it and resubmit it.

The exponential distribution is a continuous probability distribution with probability density definded by: f(x) = ke-kx for x â‰¥ 0 and f(x) = 0 otherwise.

There is the normal probability density function (pdf) which is given in the attached link. The normal probability cumulative distribution function (cdf) is used to calculate probabilities, and there is no closed form equation for this. Many statistical programs have the cdf built in. Some references are given at the end of the link to find approximate cdf. The cdf, is usually written F(x) and the pdf f(x). F(x) is the integral of f(x) from minus infinity to x.

If f(x, y) is the joint probability distribution function of two random variables, X and Y, then the sum (or integral) of f(x, y) over all possible values of y is the marginal probability function of x. The definition can be extended analogously to joint and marginal distribution functions of more than 2 variables.

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.

I think you left off some important information. Perhaps you can supply this information, to obtain assistance. To calculate the probability or the chance of occurrence between two values, we calculate: Pr{a < X < b} = F(b) - F(a) where F(x) = cumulative probability distribution. The distribution requires certain known parameters. In the case of the Normal distribution, the mean and standard deviation are parameters. In your particular case, a = 20 and b = 28.

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.

f(x) = ...f is the name of the function, and x is the variable. I guess you could say x is the root of the function, because it is what the function relies on.

Suppose you have two random variables, X and Y and their joint probability distribution function is f(x, y) over some appropriate domain. Then the marginal probability distribution of X, is the integral or sum of f(x, y) calculated over all possible values of Y.