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प्रश्न
Find the mean, variance and standard deviation for the data:
227, 235, 255, 269, 292, 299, 312, 321, 333, 348.
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उत्तर
227,235,255,269,292,299,312,321,333,348,
\[\text{ Mean } = \frac{227 + 235 + 255 + 269 + 292 + 299 + 312 + 321 + 333 + 348}{10}\]
\[ = \frac{2891}{10} = 289 . 1\]
|
\[x_i\]
|
\[\left( x_i - X \right) = \left( x_i - 289 . 1 \right)\]
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\[\left( x_i - X \right)^2\]
|
|---|---|---|
| 227 | − 62.1 | 3856.41 |
| 235 | − 54.1 | 2926.81 |
| 255 | − 34.1 | 1162.81 |
| 269 | − 20.1 | 404.01 |
| 292 | 2.9 | 8.41 |
| 299 | 9.9 | 98.01 |
| 312 | 22.9 | 524.41 |
| 321 | 31.9 | 1017.61 |
| 333 | 43.9 | 1927.21 |
| 348 | 58.9 | 3469.21 |
|
\[\sum^{10}_{i = 1} \left( x_i - \bar{x} \right)^2 = 15394 . 9\]
|
n = 10
\[n = 10\]
\[ \therefore \text{ Variance } \left( X \right) = \frac{\sum^{10}_{i = 1} \left( x_i - \bar{X} \right)^2}{n} \]
\[ = \frac{15394 . 9}{10} \]
\[ = 1539 . 49\]
\[\text{ Standard deviation } = \sqrt{\text{ Variance } \left( X \right)} \]
\[ = \sqrt{1539 . 49} \]
\[ = 39 . 24\]
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