Question
Download Solution PDFयदि 10 प्रेक्षणों का योग 12 है और उनके वर्गों का योग 18 है, तो मानक विचलन है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFमानक विचलन = \(\sqrt{{{(x_i - \bar x)}^2} \over n} = \sqrt {Variance} \)
इसलिए, प्रसरण V(X) = \({1 \over n} {{(x_i - \bar x)}^2} \)
या, \(\rm v(x)=\frac{1}{n}\left[\Sigma(x_i^2-2\times x_i \times \bar x+\bar x^2)\right]\)
\(\rm =\frac{\Sigma x_i^2}{n}-2 \bar x\left(\frac{\Sigma x_i}{n}\right)+\frac{\Sigma \bar x^2}{n}\)
\(=\rm \frac{\Sigma x_i^2}{n}-2\bar x^2+\frac{n.\bar x^2}{n}\)
\(\rm = \frac{\Sigma x_i^2}{n}-2\bar x^2+\bar x^2\)
\(\rm V(x)=\frac{\Sigma x_i^2}{n}-\bar x^2\)
दिया गया है, 10 प्रेक्षणों का योग 12 है और उनके वर्गों का योग 18 है।
इस प्रकार, n = 10, \(\sum x_i = 12\), \(\sum x_i^2 = 18\)
इसलिए, माध्य \(\bar x = 1.2\)
इस सूचना को प्रसरण V (X) के समीकरण में रखने पर, हमें प्राप्त होता है
\(\rm V(x)=\frac{\Sigma x_i^2}{n}-\bar x^2\)
\(=\frac{1}{10}\times 18-(1.2)^2\)
= 1.8 - 1.44
=0.36
इसलिए, अभीष्ट मानक विचलन =
\(\sqrt {Variance} = \sqrt{(0.36)} = 0.6 = {3 \over 5}\)
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