Find the P60 and the median of the given data
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Following are the calculation of the Variance, standard deviation, and mean:
Given:
[tex]Class \ interval \ \ \ \ \ \ \ \ \ \ \ F \\\\30-34 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \\\\35-39 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11\\\\40-44 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7\\\\45-49 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22\\\\[/tex]
[tex]Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 43[/tex]
Find:
[tex]Variance=?\\\\Standard\ Deviation=?\\\\Sample \ Mean=?[/tex]
Solution:
[tex]Class \ interval \ \ \ \ \ \ \ \ \ X \ \ \ \ \ \ \ \ \ \ \ \ \ F \ \ \ \ \ \ \ \ \ \ \ \ \ FX \ \ \ \ \ \ \ \ \ \ \ \ X^2 \ \ \ \ \ \ \ \ \ \ \ \ FX^2 \\\\30-34 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \ \ \ \ 96 \ \ \ \ \ \ \ \ \ \ \ \ 1024 \ \ \ \ \ \ \ \ \ \ \ \ 3072\\[/tex]
[tex]35-39\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 37 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 11\ \ \ \ \ \ \ \ \ \ \ \ 407 \ \ \ \ \ \ \ \ 1369 \ \ \ \ \ \ \ \ 15059 \\40-44 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 42\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 7 \ \ \ \ \ \ \ \ \ \ \ \ \ 294 \ \ \ \ \ \ \ \ \ \ 1764 \ \ \ \ \ \ \ \ \ 12,348\\45-49\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 47 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22 \ \ \ \ \ \ \ \ \ \ \ 1034 \ \ \ \ \ \ \ \ \ \ \ 2209 \ \ \ \ \ \ \ \ \ \ 48,598[/tex]
[tex]Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 43 \ \ \ \ \ \ \ \ \ \ 1831\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 79077[/tex]
[tex]\to \bold{Variance \ \sigma^2 = \frac{ \Sigma Fx^2}{\Sigma X}-\bar{X^2}}[/tex]
[tex]\bold{ = \frac{ 79077}{43} - (\frac{1831}{43})^2}\\\\\bold{ = 1839 - (1813.17)}\\\\\bold{ = 25.83}[/tex]
[tex]\to \bold{Standard\ Deviation =\sqrt{\sigma^2}}[/tex] [tex]\bold{=\sqrt{25.83} \approx 5.08}[/tex]
[tex]\to \bold{Sample\ Mean\ \bar{X}=\frac{\Sigma FX}{\Sigma F} =\frac{1831}{43}= 42.581= 42.59\approx 42.5}[/tex]
Learn more:
brainly.com/question/2141702