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Universidad politécnica territorial del Estado Lara Andrés Eloy
Blanco PNF de distribución y logística
GUERRERO, NICOLE. 28493029 DYL 2200
Solución:
Para hallarla mediade muestra,debemosprimerodesarrollarlatablade frecuenciade la
situaciónplanteada
Estatura (metros) xi fi (xi)(fi) Fi
1,65-1,69 1,67 6 10,02 6
1,70-1,74 1,72 12 20,64 18
1,75-1,79 1,77 30 53,1 48
1,80-1,84 1,82 22 40,04 70
1,85-1,89 1,87 8 14,96 78
1,90-1,94 1,92 2 3,84 80
TOTAL 80 142,6
 La mediamuestral se calculacomosigue:
𝑥̅ = ∑
𝑥𝑖𝑓𝑖
80
6
𝑖=1
=
142,6
80
= 1,78
La estatura promediode los80 estudiantesesde 1,78 metros
 La mediana,se calculaconla expresión: 𝑀𝑒 = 𝐿𝑖 + (
𝑛
2
− 𝐹𝑖−1
𝑓𝑖
). 𝑎
El intervalodonde se encuentralamedianaesel primerintervalo enel cual se cumple:
𝑛
2
=
80
2
= 40 ≤ 𝐹𝑖
40 ≤ 48
El 3er intervalo:
𝑖 = 3
𝑎 = 0,05
𝐹2 = 18
𝑓3 = 30
𝐿3 = 1,745
𝑀𝑒 = 1,745 + (
40 − 18
30
). 0,05 = 1,781
El valor que separa a los estudiantesendos grupos de igual cantidad es de 1,781
 Para calcularla moda,vamosa estudiardoscasos(a y b):
Caso a: El intervalode mayorfrecuenciaesel 3ero
𝑴𝒐 = 𝑳𝒊 + (
𝒅𝒊
𝒅𝟏+ 𝒅𝟐
). 𝒂
f3=30
a=0,05
Caso b: El intervalode mayorfrecuenciaesel 3ero
f(3+1)=f4=22
f(3-1)=f2=12
a=0,05
Solución:Expresamosenlatablade frecuencialafrecuenciaacumuladaparapodercalcularla
mediana
Preciode ventas($) fi Fi
0-5 310 310
5-10 430 740
10-15 480 1220
15-20 520 1740
20-25 500 2240
25-30 490 2730
30-35 420 3150
35-40 370 3520
40-45 260 3780
45-50 110 3890
TOTAL 3890
La mediana,se calculaconla expresión:
El intervalodonde se encuentralamedianaesel primerintervaloenel cual:
El 5to intervalo:
El valor que separa a las acciones endos grupos de igual cantidad esde 22,05$

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Actividad 3 nicole guerrero

  • 1. Universidad politécnica territorial del Estado Lara Andrés Eloy Blanco PNF de distribución y logística GUERRERO, NICOLE. 28493029 DYL 2200 Solución: Para hallarla mediade muestra,debemosprimerodesarrollarlatablade frecuenciade la situaciónplanteada Estatura (metros) xi fi (xi)(fi) Fi 1,65-1,69 1,67 6 10,02 6 1,70-1,74 1,72 12 20,64 18 1,75-1,79 1,77 30 53,1 48 1,80-1,84 1,82 22 40,04 70 1,85-1,89 1,87 8 14,96 78 1,90-1,94 1,92 2 3,84 80 TOTAL 80 142,6  La mediamuestral se calculacomosigue: 𝑥̅ = ∑ 𝑥𝑖𝑓𝑖 80 6 𝑖=1 = 142,6 80 = 1,78 La estatura promediode los80 estudiantesesde 1,78 metros
  • 2.  La mediana,se calculaconla expresión: 𝑀𝑒 = 𝐿𝑖 + ( 𝑛 2 − 𝐹𝑖−1 𝑓𝑖 ). 𝑎 El intervalodonde se encuentralamedianaesel primerintervalo enel cual se cumple: 𝑛 2 = 80 2 = 40 ≤ 𝐹𝑖 40 ≤ 48 El 3er intervalo: 𝑖 = 3 𝑎 = 0,05 𝐹2 = 18 𝑓3 = 30 𝐿3 = 1,745 𝑀𝑒 = 1,745 + ( 40 − 18 30 ). 0,05 = 1,781 El valor que separa a los estudiantesendos grupos de igual cantidad es de 1,781  Para calcularla moda,vamosa estudiardoscasos(a y b): Caso a: El intervalode mayorfrecuenciaesel 3ero 𝑴𝒐 = 𝑳𝒊 + ( 𝒅𝒊 𝒅𝟏+ 𝒅𝟐 ). 𝒂 f3=30 a=0,05 Caso b: El intervalode mayorfrecuenciaesel 3ero
  • 3. f(3+1)=f4=22 f(3-1)=f2=12 a=0,05 Solución:Expresamosenlatablade frecuencialafrecuenciaacumuladaparapodercalcularla mediana Preciode ventas($) fi Fi 0-5 310 310 5-10 430 740 10-15 480 1220 15-20 520 1740 20-25 500 2240 25-30 490 2730 30-35 420 3150 35-40 370 3520 40-45 260 3780
  • 4. 45-50 110 3890 TOTAL 3890 La mediana,se calculaconla expresión: El intervalodonde se encuentralamedianaesel primerintervaloenel cual: El 5to intervalo: El valor que separa a las acciones endos grupos de igual cantidad esde 22,05$