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Find the sample standard deviation for the datax 1 2 3 4 5 6 7 8 9 10f 384 208 98 56 28 12 8 2 3 1Round to three decimals.

Question

Find the sample standard deviation for the datax 1 2 3 4 5 6 7 8 9 10f 384 208 98 56 28 12 8 2 3 1Round to three decimals.

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Solution 1

Para encontrar la desviación estándar de la muestra para los datos proporcionados, sigue estos pasos:

  1. Calcular la media (promedio) de los datos ponderados:

    Primero, necesitamos calcular la media de los datos. La fórmula para la media ponderada es:

    xˉ=(xifi)fi \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}

    Donde xix_i son los valores de los datos y fif_i son las frecuencias correspondientes.

    xˉ=(1384)+(2208)+(398)+(456)+(528)+(612)+(78)+(82)+(93)+(101)384+208+98+56+28+12+8+2+3+1 \bar{x} = \frac{(1 \cdot 384) + (2 \cdot 208) + (3 \cdot 98) + (4 \cdot 56) + (5 \cdot 28) + (6 \cdot 12) + (7 \cdot 8) + (8 \cdot 2) + (9 \cdot 3) + (10 \cdot 1)}{384 + 208 + 98 + 56 + 28 + 12 + 8 + 2 + 3 + 1}

    Calculamos el numerador:

    (1384)+(2208)+(398)+(456)+(528)+(612)+(78)+(82)+(93)+(101)=384+416+294+224+140+72+56+16+27+10=1639 (1 \cdot 384) + (2 \cdot 208) + (3 \cdot 98) + (4 \cdot 56) + (5 \cdot 28) + (6 \cdot 12) + (7 \cdot 8) + (8 \cdot 2) + (9 \cdot 3) + (10 \cdot 1) = 384 + 416 + 294 + 224 + 140 + 72 + 56 + 16 + 27 + 10 = 1639

    Calculamos el denominador:

    384+208+98+56+28+12+8+2+3+1=800 384 + 208 + 98 + 56 + 28 + 12 + 8 + 2 + 3 + 1 = 800

    Entonces, la media es:

    xˉ=1639800=2.04875 \bar{x} = \frac{1639}{800} = 2.04875

  2. Calcular la suma de los cuadrados de las diferencias de cada valor con la media, ponderadas por sus frecuencias:

    La fórmula para la varianza ponderada es:

    s2=fi(xixˉ)2fi1 s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i - 1}

    Calculamos cada término (xixˉ)2(x_i - \bar{x})^2:

    (12.04875)2=1.099 (1 - 2.04875)^2 = 1.099 (22.04875)2=0.002 (2 - 2.04875)^2 = 0.002 (32.04875)2=0.903 (3 - 2.04875)^2 = 0.903 (42.04875)2=3.808 (4 - 2.04875)^2 = 3.808 (52.04875)2=8.713 (5 - 2.04875)^2 = 8.713 (62.04875)2=15.618 (6 - 2.04875)^2 = 15.618 (72.04875)2=24.523 (7 - 2.04875)^2 = 24.523 (82.04875)2=35.428 (8 - 2.04875)^2 = 35.428 (92.04875)2=48.333 (9 - 2.04875)^2 = 48.333 (102.04875)2=63.238 (10 - 2.04875)^2 = 63.238

    Ahora multiplicamos cada uno por su frecuencia y sumamos:

    fi(xixˉ)2=(3841.099)+(2080.002)+(980.903)+(563.808)+(288.713)+(1215.618)+(824.523)+(235.428)+(348.333)+(163.238) \sum f_i (x_i - \bar{x})^2 = (384 \cdot 1.099) + (208 \cdot 0.002) + (98 \cdot 0.903) + (56 \cdot 3.808) + (28 \cdot 8.713) + (12 \cdot 15.618) + (8 \cdot 24.523) + (2 \cdot 35.428) + (3 \cdot 48.333) + (1 \cdot 63.238)

    =422.016+0.416+88.494+213.248+243.964+187.416+196.184+70.856+144.999+63.238=1630.831 = 422.016 + 0.416 + 88.494 + 213.248 + 243.964 + 187.416 + 196.184 + 70.856 + 144.999 + 63.238 = 1630.831

  3. Calcular la varianza:

    s2=1630.8318001=1630.831799=2.041 s^2 = \frac{1630.831}{800 - 1} = \frac{1630.831}{799} = 2.041

  4. Calcular la desviación estándar:

    s=2.041=1.429 s = \sqrt{2.041} = 1.429

Por lo tanto, la desviación estándar de la muestra, redondeada a tres decimales, es 1.4291.429.

This problem has been solved

Solution 2

Para encontrar la desviación estándar de la muestra para los datos proporcionados, sigue estos pasos:

  1. Calcular la media (promedio) de los datos ponderados:

    Primero, necesitamos calcular la media de los datos. La media ponderada se calcula usando la fórmula:

    xˉ=(xifi)fi \bar{x} = \frac{\sum (x_i \cdot f_i)}{\sum f_i}

    Donde xi x_i son los valores de los datos y fi f_i son las frecuencias correspondientes.

    [ \bar{x} = \frac{(1 \cdot 384) + (2 \cdot 208) + (3 \cdot 98) + (4 \cdot 56) + (5 \cdot 28) + (6 \cdot 12) + (7 \

This problem has been solved

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