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| 1. |
Find mean diameter of following circle 33-36=15,37-40=17,41-44=21,45-48=22,49-52=25. |
| Answer» \tDiametersClassesMid vales xifi{tex}u = \\frac{{x - 42.5}}{4}{/tex}fufu233-3632.5-36.534.515- 2- 306037-4036.5-40.538.517- 1- 171741-4440.5-44.542.52100045-4844.5-52.546.5221222249-5248.5-52.550.525250100\xa0\xa0\xa0100\xa025199\tMean\xa0{tex}(\\bar x) = A + \\frac{{\\Sigma fu}}{N} \\times h{/tex}{tex} = 42.5 + \\frac{{25}}{{100}} \\times 4{/tex}= 42.5 + 1 = 43.5 mmStandard deviation\xa0{tex}(\\sigma ) = \\frac{h}{N}\\sqrt {N\\Sigma f{u^2} - {{(\\Sigma fu)}^2}}{/tex}{tex} = \\frac{4}{{100}}\\sqrt {100 \\times 199 - {{(25)}^2}} = \\frac{1}{{25}}\\sqrt {19275}{/tex}{tex} = \\frac{1}{{25}} \\times 138.83{/tex} = 5.55 | |