To calculate the variance in this case, we need to use the variance for a discrete variable and its probability.
The equation we need to use in this case is:

Where x_i is each value in the first row and P(x_i) is each value in the second row.
First, let's calculate each square difference:

Now, we need to multiply each for its corresponding P(x):

Finally, we sum them all to get the variance:

So, the variance to one decimal place is 2.4.