Math  /  Data & Statistics

QuestionTaller: formativo. Est 215 sem 22024 Resuelta los siguientes problemas aplicando lo aprendido para elaborar una tabla de distribución de frecuencias. Y además realice el calculo de la media y la desviación típica.
Problema 1. En un problema similar al del ejercicio 11, es necesario garantizar que la resistencia mínima que tiene un envase de plástico en posición vertical sea de 20 kg. Para evaluar esto se han obtenido los siguientes datos mediante pruebas destructivas: 28.326.826.626.528.124.827.426.229.428.624.925.230.427.727.026.128.126.928.027.625.629.527.627.326.227.727.225.926.528.326.529.123.729.726.829.528.426.328.128.727.025.526.927.227.625.528.327.428.825.025.327.725.228.627.928.7\begin{array}{l} 28.3 \quad 26.8 \quad 26.6 \quad 26.5 \quad 28.124 .827 .426 .2 \quad 29.428 .624 .925 .230 .427 .727 .026 .1 \\ 28.126 .928 .027 .625 .629 .527 .627 .326 .227 .727 .225 .926 .528 .326 .529 .1 \\ 23.729 .726 .829 .528 .426 .328 .128 .727 .025 .526 .927 .227 .625 .528 .327 .4 \\ 28.825 .025 .3 \quad 27.725 .228 .627 .928 .7 \end{array}

Studdy Solution
Calculate the standard deviation (σ\sigma):
σ=σ2=2.4161.555 \sigma = \sqrt{\sigma^2} = \sqrt{2.416} \approx 1.555
The mean resistance value is approximately 27.7136 27.7136 kg, and the standard deviation is approximately 1.555 1.555 kg.

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