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Sagot :
¡Claro! Vamos a completar la tabla paso a paso.
Tenemos la siguiente información:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline & f_i & F_i & h_i & H_i \\ \hline & & & & 0.1 \\ \hline & & & & 0.4 \\ \hline & 24 & & 0.20 & \\ \hline & 30 & & 0.25 & \\ \hline \text{TOTAL} & N = & & 0.15 & \\ \hline \end{array} \][/tex]
### Paso 1: Llenar [tex]\( H_i \)[/tex]
Sabemos que [tex]\( H_i \)[/tex] es la frecuencia relativa acumulada.
1. [tex]\( H_i[2] \)[/tex] = [tex]\( H_i[1] + h_i[2] \)[/tex] = [tex]\( 0.4 + 0.2 = 0.6 \)[/tex]
2. [tex]\( H_i[3] \)[/tex] = [tex]\( H_i[2] + h_i[3] \)[/tex] = [tex]\( 0.6 + 0.25 = 0.85 \)[/tex]
Actualizando [tex]\( H_i \)[/tex]:
[tex]\[ H_i = [0.1, 0.4, 0.6, 0.85] \][/tex]
### Paso 2: Calcular [tex]\( h_i \)[/tex] faltante
Sabemos que la suma de todas las frecuencias relativas ([tex]\( h_i \)[/tex]) debe ser 1.
3. [tex]\( h_i \)[/tex] faltante = [tex]\( 1.0 - (0.2 + 0.25) = 0.55 \)[/tex]
Entonces:
[tex]\[ h_i = [0.1, 0.4, 0.2, 0.25] \][/tex]
### Paso 3: Calcular [tex]\( N \)[/tex] (total de frecuencias)
La suma de las frecuencias relativas ([tex]\( h_i \)[/tex]) debería equivaler al total de frecuencias.
4. [tex]\( N = 24 / 0.2 = 120 \)[/tex]
### Paso 4: Llenar [tex]\( f_i \)[/tex]
5. [tex]\( f_i[0] = N \times h_i[0] = 120 \times 0.1 = 12 \)[/tex]
6. [tex]\( f_i[1] = N \times h_i[1] = 120 \times 0.4 = 48 \)[/tex]
Actualizando [tex]\( f_i \)[/tex]:
[tex]\[ f_i = [12, 48, 24, 30] \][/tex]
### Paso 5: Calcular [tex]\( F_i \)[/tex]
7. [tex]\( F_i[0] = f_i[0] = 12 \)[/tex]
8. [tex]\( F_i[1] = f_i[0] + f_i[1] = 12 + 48 = 60 \)[/tex]
9. [tex]\( F_i[2] = F_i[1] + f_i[2] = 60 + 24 = 84 \)[/tex]
10. [tex]\( F_i[3] = F_i[2] + f_i[3] = 84 + 30 = 114 \)[/tex]
Actualizando [tex]\( F_i \)[/tex]:
[tex]\[ F_i = [12, 60, 84, 114] \][/tex]
Ahora la tabla completa es:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline & f_i & F_i & h_i & H_i \\ \hline & 12 & 12 & 0.1 & 0.1 \\ \hline & 48 & 60 & 0.4 & 0.4 \\ \hline & 24 & 84 & 0.20 & 0.6 \\ \hline & 30 & 114 & 0.25 & 0.85 \\ \hline \text{TOTAL} & 114 & & 1.0 & \\ \hline \end{array} \][/tex]
Tenemos la siguiente información:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline & f_i & F_i & h_i & H_i \\ \hline & & & & 0.1 \\ \hline & & & & 0.4 \\ \hline & 24 & & 0.20 & \\ \hline & 30 & & 0.25 & \\ \hline \text{TOTAL} & N = & & 0.15 & \\ \hline \end{array} \][/tex]
### Paso 1: Llenar [tex]\( H_i \)[/tex]
Sabemos que [tex]\( H_i \)[/tex] es la frecuencia relativa acumulada.
1. [tex]\( H_i[2] \)[/tex] = [tex]\( H_i[1] + h_i[2] \)[/tex] = [tex]\( 0.4 + 0.2 = 0.6 \)[/tex]
2. [tex]\( H_i[3] \)[/tex] = [tex]\( H_i[2] + h_i[3] \)[/tex] = [tex]\( 0.6 + 0.25 = 0.85 \)[/tex]
Actualizando [tex]\( H_i \)[/tex]:
[tex]\[ H_i = [0.1, 0.4, 0.6, 0.85] \][/tex]
### Paso 2: Calcular [tex]\( h_i \)[/tex] faltante
Sabemos que la suma de todas las frecuencias relativas ([tex]\( h_i \)[/tex]) debe ser 1.
3. [tex]\( h_i \)[/tex] faltante = [tex]\( 1.0 - (0.2 + 0.25) = 0.55 \)[/tex]
Entonces:
[tex]\[ h_i = [0.1, 0.4, 0.2, 0.25] \][/tex]
### Paso 3: Calcular [tex]\( N \)[/tex] (total de frecuencias)
La suma de las frecuencias relativas ([tex]\( h_i \)[/tex]) debería equivaler al total de frecuencias.
4. [tex]\( N = 24 / 0.2 = 120 \)[/tex]
### Paso 4: Llenar [tex]\( f_i \)[/tex]
5. [tex]\( f_i[0] = N \times h_i[0] = 120 \times 0.1 = 12 \)[/tex]
6. [tex]\( f_i[1] = N \times h_i[1] = 120 \times 0.4 = 48 \)[/tex]
Actualizando [tex]\( f_i \)[/tex]:
[tex]\[ f_i = [12, 48, 24, 30] \][/tex]
### Paso 5: Calcular [tex]\( F_i \)[/tex]
7. [tex]\( F_i[0] = f_i[0] = 12 \)[/tex]
8. [tex]\( F_i[1] = f_i[0] + f_i[1] = 12 + 48 = 60 \)[/tex]
9. [tex]\( F_i[2] = F_i[1] + f_i[2] = 60 + 24 = 84 \)[/tex]
10. [tex]\( F_i[3] = F_i[2] + f_i[3] = 84 + 30 = 114 \)[/tex]
Actualizando [tex]\( F_i \)[/tex]:
[tex]\[ F_i = [12, 60, 84, 114] \][/tex]
Ahora la tabla completa es:
[tex]\[ \begin{array}{|l|l|l|l|l|} \hline & f_i & F_i & h_i & H_i \\ \hline & 12 & 12 & 0.1 & 0.1 \\ \hline & 48 & 60 & 0.4 & 0.4 \\ \hline & 24 & 84 & 0.20 & 0.6 \\ \hline & 30 & 114 & 0.25 & 0.85 \\ \hline \text{TOTAL} & 114 & & 1.0 & \\ \hline \end{array} \][/tex]
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