Question
Download Solution PDFFor the series 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, the minimum possible value of \(\rm \Sigma_{i=1}^n(x_i-A)^2\) can be attained at:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFTo find the value of A that minimizes the sum \(\rm \Sigma_{i=1}^n(x_i-A)^2\), we need to find the value of A that minimizes the squared differences from the elements of series.
For the given series x1, x2, ...., xn, the value of A that maximizes \(\rm \Sigma_{i=1}^n(x_i-A)^2\), is the mean (average) of the series.
The given series is:
\(\frac{1+2+3+4+5+6+7+8+9+10}{10}=\frac{55}{10}=5.5\)
So the value of A that minimizes \(\rm \Sigma_{i=1}^n(x_i-A)^2\)= 5.5
Last updated on Jul 17, 2025
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