Outlier and correlation 
Hi, I have a question.
The scatter plot doesn't show any type of correlation and there is an outlier. 
If the outlier was to be removed, would the correlation:


*

*Increase dramatically

*Increase slightly

*Decrease slightly

*Remain unchanged


Why?
I think it would remain unchanged because there is no correlation. 
The correlation (r) is listed as 0.20 if needed. 
Thank you for your help in advance!
 A: Visualizing the addition of an outlier at upper-right:
49 Original points. Start with $n = 49$ points simulated according to an uncorrelated $(\rho=0)$ bivariate normal distribution. The sample correlation $(r = -0.0096)$ is very nearly 0. [Computations in R.]
set.seed(1006)
x = rnorm(49); y = rnorm(49)
cov(x,y);  sd(x);  sd(y)
[1] -0.01219304
[1] 1.116661
[1] 1.141112
cov(x,y)/(sd(x)*sd(y)); cor(x,y)
[1] -0.009568911
[1] -0.009568911

Plot the 49 points.
par(mfrow=c(1,2))
 plot(x,y, ylim=c(-3,6), xlim=c(-3,6), pch=19)
  abline(v=0, col="green2"); abline(h=0, col="green2")


One additional point. Add an outlier at $(6,6)$ and make a new plot that includes this 50th point (colored blue).
  x = c(x,6); y = c(y,6)
  plot(x,y, ylim=c(-3,6), xlim=c(-3,6), pch=19)
    points(6,6, pch=19, col="blue")
    abline(v=0, col="green2"); abline(h=0, col="green2")
par(mfrow=c(1,1))


Re-compute the correlation of all fifty points; the new correlation is
$r = 0.3693.$
cov(x,y);  sd(x);  sd(y)
[1] 0.7379323
[1] 1.397288
[1] 1.429992
cov(x,y)/(sd(x)*sd(y));  cor(x,y)
[1] 0.3693151
[1] 0.3693151

Formulas. The sample covariance is $S_{XY} = \frac{1}{n-1}\sum_{i=1}^{50}(X_i-\bar X)(Y_i - \bar Y).$ Products corresponding to points in the first and third quadrants
are positive and products corresponding to points in the second and fourth
quadrants are negative. 
The 50th point, a far outlier in the first quadrant, added about $6 \times 6 = 36$ to the sum in the
covariance, which was about 0 before (with 49 relatively small counter-balancing positive and negative products). The correlation $r = \frac{S_{XY}}{S_XS_Y}$
increased considerably because the covariance in the numerator is now
considerably larger.
A: Think about the following example: you have a lot of equally spaced points around unit circle centered at the origin. There is really no correlation at all. Then, you add another point to the graph,  e.g. $(10,10)$. Suddenly, it is more possible to do a line fit to your data, (btw. correlation can even go up to very high levels, e.g. $0.9$) which in turn means higher Pearson correlation. Your data resembles this example. So, your Pearson correlation will decrease. But, the dramatical and slight definitions here are obscure. However, based on this figure, it is possible to have your value decrease around $0.2$, which is noticeable in my opinion.
