Changes
From Karnataka Open Educational Resources
381 bytes added
, 09:56, 30 October 2019
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| =Problem 1= | | =Problem 1= |
− | If n =10, = 12 and | + | If n =10, <math>\bar x </math> = 12 and<br> |
− | <math>\sum{x^2}= 1530 find the standard deviation </math> <br> | + | <math>\sum{x^2}= 1530 find the standard deviation </math> <br> |
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| =INTERPRETATION OF PROBLEM= | | =INTERPRETATION OF PROBLEM= |
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− | |128 | + | |7 |
− | |080 | + | |36 |
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| |50-54 | | |50-54 |
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| + | |9 |
| + | |9 |
| + | |27 |
| + | |- |
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| + | |n=40 |
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| + | |Σfx=1668 |
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− | | | + | |Σfd=17 |
− | |n=50
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− | |Σfx=660
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− | |Σfx²=10710 | + | |Σfd²=85 |
| |} | | |} |
| + | A=assumed average.<br> |
| + | c=4<br> |
| + | d=<math>\frac{x-A}{c}</math>=<math>\frac{32-40}{4}</math>=<math>\frac{-8}{4}=-2</math><br> |
| + | |
| + | assumed mean A=<math>\frac{\sum fx}{n}</math>=<math>\frac{1668}{40}=41.7</math><br> |
| + | |
| + | Varience σ²=[<math>\frac{\sum {fd^2}}{n}-({\frac{\sum fx}{n})^2}]c^2</math> <br> |
| + | |
| + | σ²=[<math>\frac{85}{40}-({\frac{17}{40})^2}]4^2</math> <br> |
| + | |
| + | σ²=[2.125-0.180]16<br> |
| | | |
| + | σ²=[1.945]16<br> |
| | | |
| + | σ²=31.12<br> |
| + | standard deviation, σ=<math>\sqrt{varience}</math> <br> |
| + | σ=<math>\sqrt{31.12}</math> <br> |
| + | σ=5.58 |
| | | |
− | Standard deviation σ=<math>\sqrt{\frac {\sum {fx^2}}{n}-({\frac{\sum fx}{n})^2}}</math> <br>
| + | [[Category:Statistics]] |
− | σ=<math>\sqrt{\frac{10700}{50}-({\frac{660}{50})^2}}</math> <br>
| |
− | σ=<math>\sqrt{214-174.24}</math> <br>
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− | σ=<math>\sqrt{39.96}</math> <br>
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− | σ=6.3
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