Calculate mean from cumulative percentage table Is it possible to calculate mean from cumulative percentage table below?
Scenario: Bob is a manager. He is trying to estimate the sales for the next year, and from his experience, he knows the sales should go like the table below. Now, he would like to know what's the mean of the sales.
Thank you!
Sales   | Possibility that sales below this number
 100    0%
 105    5.00%
 110    10.00%
 116    15.00%
 122    20.00%
 128    25.00%
 134    35.00%
 141    45.00%
 148    55.00%
 155    65.00%
 163    75.00%
 171    90.00%
200 100%
 A: I understand this question is about two years old, but it was recently poked by Community and so came to my attention.  I'd solve this problem using the relationship between a survival function of a non-negative random variable and its expectation. Let $X$ be a non-negative random variable (here $X$ is total sales). Then $EX = \int_0^\infty (1-F(x))\ dx$, where $F(x) = P(X \leq x)$ denotes the cumulative distribution function (CDF) of $X$. From the data you have, we can see that $P(100 \leq X \leq 200) = 1$ (implying $X$ is non-negative). So, we have
$$
\begin{aligned}
EX &= \int_{0}^{\infty} (1-F(x)) \ dx \\
&= \int_{0}^{200} (1-F(x)) \ dx \qquad \text{(since } 1 - F(x) = P(X > x) = 0 \text{ for } x \geq 200) \\ 
&= (200 - 0) - \int_{100}^{200} F(x) \ dx \\
&= 200 - \left[\int_{100}^{105} F(x) \ dx + \int_{105}^{110} F(x) \ dx + \dots + \int_{171}^{200} F(x) \ dx \right] 
\end{aligned}
$$
Since we don't know the CDF completely, all we can do is some approximation. A simple (albeit crude) approximation is given by the composite trapezoidal method. Here we approximate each of the above sub-integrals using the Trapezoidal rule, which says that the definite integral 
$$
\int_{a}^{b} F(x) \, dx \approx (b-a) \cdot \frac{F(a)+F(b)}{2}.
$$ 
So, e.g., $\int_{100}^{105} F(x) \ dx \approx (105 - 100)  [F(100) + F(105)]/2 = 5 * (0 + 0.05)/2 = 0.125$.  An R implementation of this approximation scheme is as follows:
x <- c(100, 105, 110, 116, 122, 128, 134, 
       141, 148, 155, 163, 171, 200)
Fx <- c(0, 0.05, 0.10, 0.15, 0.20, 0.25, 0.35, 
        0.45, 0.55, 0.65, 0.75, 0.90, 1)
n <- length(Fx)
EX.approx <- 200 - sum(diff(x) * (Fx[-n] + Fx[-1])/2)
EX.approx
# [1] 144.3

Another approximation will be obtained by first linearly interpolating $F$ in $[100, 200]$ based on the data we have, and then doing a numerical integration over that interpolated function. Here is an R implementation based on this interpolation approach: 
Fx_smooth <- approxfun(x = x, y = Fx, method = "linear")
EX.approx2 <- 200 - integrate(Fx_smooth, 
                              lower = 100, 
                              upper = 200)$val
EX.approx2
# [1] 144.2997

Edits: Grammar. 
A: So, given a probability distribution p(x), the mean is the integral of x*p(x) over the relevant domain (i.e. where p(x) is non-zero). What you have there isn't a probability distribution, though, it's an integrated probability distribution (which I'll call "h(x)" in this post) evaluated at a finite set of points. So what we know, for example, is the integral of p(x) from -inf to 100 is 0, the integral of p(x) from 100 to 105 is 5%, the integral of p(x) from 100 to 110 is 10%, etc. If we were given the function h(x) that gave all of these percentage values, for every x, then we can in fact recover p(x) by taking the derivative of h(x).  With h(x) evaluated at only a finite set of points, there is no way to reconstruct p(x), and therefore no way to obtain the mean.
At best, you can give an approximation of the mean, by making a few additional assumptions. For example, you can assume the probability distribution p(x) consists of segments of horizontal lines (a collection of uniform distributions over the domain). In this, case you split your domain up into 12 segments, you take the midpoint of the segment and multiply it by the probability in that segment (this is performing the integral for a bunch of horizontal lines). So concretely you would do:
102.5*5%+107.5*5%+113*5%+... and get an estimate that way. But this is just a guess at the mean, depending on how the probability distribution is actually shaped in between each interval, the mean could change.  
