The linear regression for a given data set has the form

where the values a and b can be solved using the equation

Based on the given data set, we have n equals 5. We will solve for the values of the summation first. We have the following

Using these values to compute for the values of a and b, we get

Take note that the problem wants us to reduce the numbers to the nearest tenth. Hence, the linear regression for the given data set is written as
