The linear correlation coefficient is 
Explanation:
It is given that
,
,
,
and 
Also, the random sample is 
The formula to determine the correlation coefficient is given by
![$r=\frac{n\left(\sum x y\right)-\left(\sum x\right)(\Sigma y)}{\sqrt{\left[n \sum x^{2}-\left(\sum x\right)^{2}\right]\left[n \Sigma y^{2}-(\Sigma y)^{2}\right]}}$](https://tex.z-dn.net/?f=%24r%3D%5Cfrac%7Bn%5Cleft%28%5Csum%20x%20y%5Cright%29-%5Cleft%28%5Csum%20x%5Cright%29%28%5CSigma%20y%29%7D%7B%5Csqrt%7B%5Cleft%5Bn%20%5Csum%20x%5E%7B2%7D-%5Cleft%28%5Csum%20x%5Cright%29%5E%7B2%7D%5Cright%5D%5Cleft%5Bn%20%5CSigma%20y%5E%7B2%7D-%28%5CSigma%20y%29%5E%7B2%7D%5Cright%5D%7D%7D%24)
Substituting the values in the formula, we have,
![$r=\frac{10(1676)-(108)(138)}{\sqrt{\left[10(1249)-(108)^{2}\right]\left[10(2280)-(138)^{2}\right]}}$](https://tex.z-dn.net/?f=%24r%3D%5Cfrac%7B10%281676%29-%28108%29%28138%29%7D%7B%5Csqrt%7B%5Cleft%5B10%281249%29-%28108%29%5E%7B2%7D%5Cright%5D%5Cleft%5B10%282280%29-%28138%29%5E%7B2%7D%5Cright%5D%7D%7D%24)
Simplifying the values, we get,
![$r=\frac{16760-14904}{\sqrt{\left[12490-11664\right]\left[22800-19044\right]}}$](https://tex.z-dn.net/?f=%24r%3D%5Cfrac%7B16760-14904%7D%7B%5Csqrt%7B%5Cleft%5B12490-11664%5Cright%5D%5Cleft%5B22800-19044%5Cright%5D%7D%7D%24)
Subtracting the values in both numerator and denominator, we have,
![$r=\frac{1856}{\sqrt{\left[826\right]\left[3756\right]}}$](https://tex.z-dn.net/?f=%24r%3D%5Cfrac%7B1856%7D%7B%5Csqrt%7B%5Cleft%5B826%5Cright%5D%5Cleft%5B3756%5Cright%5D%7D%7D%24)
Multiplying the denominator,

Simplifying, we have,

Dividing, we get,

Thus, the linear correlation coefficient is 