This problem was a great refresher on calculating Pearson’s correlation coefficient manually. The value (0.145) shows a very weak positive correlation between Physics and History scores, meaning this site subjects don’t strongly influence each other.

“Interesting problem! Karl Pearson’s correlation really shows how data can uncover hidden relationships. I recently realized how similar concepts apply even outside of academics. For example, in home improvement projects like bathroom remodeling, we often analyze the correlation between budget, material quality, and customer satisfaction. By looking at patterns, we can predict which design choices will give the best long-term results.it’s fascinating how statistics tie into real-world projects too!

The tricky part with this one is keeping the calculation clean because the data has repeats. Once you plug it into the Pearson formula, the correlation actually comes out slightly negative, around -0.144. It shows that higher Physics scores don’t really align with higher History scores in this set.

Enter your code here. Read input from STDIN. Print output to STDOUT

I have submitted this code still getting compilation error physics_marks = [15,12,8,8,7,7,7,6,5,3] history_score = [10,25,17,11,13,17,20,13,9,15]

calculate correlation coefficient

x_avg = sum(physics_marks)/10 y_avg = sum(history_score)/10 corr_coeff_num = sum([(physics_marks[i] - x_avg)*(history_score[i]-y_avg) for i in range(10)]) corr_coeff_deno = (sum([(physics_marks[i]**2)for i in range(10)])*sum([(history_score[i]**2)for i in range(10)]))**0.5 correlation_coefficient = corr_coeff_num/corr_coeff_deno print(f"{correlation_coefficient:.3f}")

physics_marks = [15,12,8,8,7,7,7,6,5,3] history_score = [10,25,17,11,13,17,20,13,9,15]

calculate correlation coefficient

correlation_coefficient = np.corrcoef(physics_marks,history_score)[0,1]

x_avg = sum(physics_marks)/10 y_avg = sum(history_score)/10 corr_coeff_num = sum([(physics_marks[i] - x_avg)*(history_score[i]-y_avg) for i in range(10)]) corr_coeff_deno = (sum([(physics_marks[i]**2)for i in range(10)])*sum([(history_score[i]**2)for i in range(10)]))**0.5 correlation_coefficient = corr_coeff_num/corr_coeff_deno print(f"{correlation_coefficient:.3f}")