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In most undergrad classes, E(X) is taught to be the "expected gain" by playing a game of bets.

For instance: If we consider a Bernoulli distribution with $p=0.3$ (Winning) and if you win, you get \$10 and you lose, you pay \$3, the E(X) = 10 x 0.3 - 3 x 0.7 = 3 - 2.1= \$0.9.

Is there a way this intuition carries forth to a continuous distribution?

I understand that E(X) can be considered are centre of mass for a distribution but I don't seem to understand what it means intuitively to say that a mean of a continuous distribution is 5.

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    The expectation is a weighted mean. In first-year calculus you probably dealt at least briefly with the idea that the mean value of a function over an interval can be obtained by integrating the function over that interval and dividing by the length of the interval. Do you see the connection between that calculation and the calculation of the mean of a finite set of numbers? That’s the connection between expectation for a continuous distribution and expectation for a discrete distribution. It may help to recall that integrals are approximated by Riemann sums; intuitively you can think of ...2012-12-12
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    ... the calculation in the continuous case as taking a weighted average of an infinite number of infinitesimal quantities.2012-12-12

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For simplicity, assume that the density function $f(x)$ is non-zero over the interval $[a,b]$. Divide the interval from $a$ to $b$ using the equally spaced points $x_0=a$, $x_1$, $x_2$, and so on up to $x_n=b$.

Think of $n$ as large. The probability that our random variable $X$ takes on values in the interval $[x_i,x_{i+1}]$ is approximately the length of the interval, namely $\dfrac{b-a}{n}$ times the density at $x_i$.

Now imagine all this probability as concentrated at the point $x_i$. So now we have a discrete distribution. Its mean, by the usual "finite" rule you quoted, is approximately $$\sum_{i=0}^{n-1}x_i \frac{b-a}{n}f(x_i).$$ Finally, let $n\to\infty$. Then the above Riemann sum approaches $$\int_a^b xf(x)\,dx.$$

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Maybe the easiest approach is to start with the uniform distribution over $(4,6)$. It seems reasonable to say that the expected value of this is $5$. The official definition is $E[X]=\int X\cdot P(X) dX$ and you can see this is satisfied. Then if you think about distributions that have stairsteps, that are constant over a couple of lengths and see that the center of mass (think of cutting the region out of a piece of graph paper) is a reasonable expected value.

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    So, that would be the mean. The median would be the value of x which divides areas equally and mode would be a stationary point. Is that correct?2012-12-12
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    @Inquest: No, the mean weights areas by the distance from the mean. It is the same as in the discrete case: the mean of $3,3,3,6$ is $4$, not $4.5$. Similarly the mean of a distribution that is $0.8$ over $(0,1)$ and $0.2$ over $(1,2)$ is $0.7$, while the area is divided evenly at $0.625$. I don't know the definition of the mode in the continuous case.2012-12-12