statistics please help!?

the following distribution table describes the probability distribution for the number of adults (among 4 randomly selected adults) who have a college degree
X P(x)
0 .0018
1 .0756
2 .2646
3 .4116
4 .2401

Find the variance for the probability distribution.

To find the variance, first you need to find the mean.

The mean is the sum of X * P(X). (* = multiply) In other words,
Then the variance is the average of the square of the distance from the mean. In other words, for each value of X, find (X - mean)^2, and multiply by the probability of that X. In other words,

variance = 0.0018 * (0 - mean)^2 + 0.0756 * (1 - mean)^2 + 0.2646 * (2 - mean)^2 .... etc

EDIT (since you've obviously already tried, I'll give my answers):

mean = 0 * 0.0018 + 1 * 0.0756 + 2 * 0.2646 + 3*0.4116 + 4*0.2401 = 2.8 (exactly, according to my PC).

variance = 0.0018 * (0 - mean)^2 + 0.0756 * (1 - mean)^2 + 0.2646 * (2 - mean)^2 + 0.4116 * (3-mean)^2 + 0.2401 * (4-mean)^2 = 0.791 (to 3 figures).

I might have made a mistake, but I don't see how the variance could be as high as 5.3. For all data points except X=0, the square of the distance from the mean is less than 4, so I don't see how the average distance squared could possibly be 5.3. On the other hand 0.0263 looks way too low.