Question

The following table contains the placement statistics for JKL institute of technology. The first column contains the CGPA or grade of the students and the second column contains the average salary in lakhs per annum. Find the average salary corresponding to 8.5 CGPA using linear interpolation.

$\begin{array}{cc}CGPA\left(X\right)& Salary\left(Y\right)\\ 5& 2.5\\ 6& 4\\ 6.5& 4.5\\ 7& 6\\ 7.5& 7.5\\ 8& 9\\ 9& 11\end{array}$

• A. 8.8 lakhs per annum
• B. 9.24 lakhs per annum
• C. 9.72 lakhs per annum
• D. 10.12 lakhs per annum

Answer 1

The correct option is C

9.72 lakhs per annum

$\begin{array}{cccccc}x& y& x-\overline{x}& (x-\overline{x}{)}^2& (y-\overline{y})& (x-\overline{x})(y-\overline{y})\\ 5& 2.5& -2& 4& -3.86& 7.72\\ 6& 4& -1& 1& -2.36& 2.36\\ 6.5& 4.5& -0.5& 0.25& -1.86& 0.93\\ 7& 6& 0& 0& -0.36& 0\\ 8& 9& 1& 1& 2.64& 2.64\\ 9& 11& 2& 4& 4.64& 9.28\\ & & & \sum =10.5& & \sum =23.5\end{array}$

$\overline{x}=\frac{\sum x}{n}=\frac{49}{7}=7$
$\overline{y}=\frac{\sum y}{n}=\frac{44.5}{7}=6.36$

$M=\frac{{\sum }_{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{{\sum }_{i=1}^n(x_i-\overline{x}{)}^2}=\frac{23.5}{10.5}=2.24$
$B=\overline{y}-M\overline{x}=6.36-2.24\times 7=-9.32$

The line of best fit is $y=2.24x-9.32$
Substituting 8.5 in the equation, we have
$y=2.24\times 8.5-9.32=9.72$

Average salary corresponding to 8.5 CGPA is 9.7 lakhs per annum