To find the regression line of y on x for the given data points (−4,1),(−1,2),(2,7), and (3,1), we start by calculating the necessary summations. These are needed for the formulas used in the method of least squares for linear regression: Number of data points n=4 Sum of x values: ∑x=−4−1+2+3=0 Sum of y values: ∑y=1+2+7+1=11 Sum of x2 values: ∑x2=(−4)2+(−1)2+(2)2+(3)2=16+1+4+9=30 Sum of xy values: ∑xy=(−4)(1)+(−1)(2)+(2)(7)+(3)(1)=−4−2+14+3=11 Using the formulas for the slope b and intercept a of the regression line y=a+bx : The slope b is given by: b=‌
nΣxy−(Σx)(Σy)
nΣx2−(Σx)2
=‌
4×11−(0)(11)
4×30−(0)2
=‌
44
120
=‌
11
30
The intercept a is given by: a=‌
Σy−bΣx
n
=‌
11−(
11
30
×0)
4
=‌
11
4
We need to find the value of 2a+15b : 2a+15b=2×‌
11
4
+15×‌
11
30
=‌
22
4
+‌
165
30
Simplify the expressions: ‌
22
4
=5.5‌‌‌ and ‌‌‌‌
165
30
=5.5 Now, add these values: 2a+15b=5.5+5.5=11 Therefore, the value of 2a+15b is 11 .