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defon
2 years ago
15

Suponga que el monto promedio de compras por cliente de una empresa es de $128.45 y una desviación de $18.26

Mathematics
1 answer:
lukranit [14]2 years ago
5 0

Answer:

Asumiendo que la distribución de las compras sea normal, tenemos:

_

x = 128.45

σ = 18.26

a)

x = 100

Normalizamos:

.......... _

z = (x - x) / σ

z = (100 - 128.45) / 18.26

z = -28.45 / 18.26

z ≅ -1.56

Buscando en la tabla de áreas bajo la curva normal tipificada (de 0 a z) obtenemos que para z = 1.56 la probabilidad es de 0.4406 = 44.06%. Esta es P(-1.56 < z < 0) = P(0 < z < 1.56).

Como preguntan P(x > 100), tenemos que

P(x > 100) = 0.5 + P(-1.56 < z < 0) = 0.5 + 0.4406 = 0.9406

La probabilidad de que un cliente compre más de $100 es del 94.06%.

=======

b)

x = 150

Normalizamos:

.......... _

z = (x - x) / σ

z = (150 - 128.45) / 18.26

z = 21.55 / 18.26

z ≅ 1.18

Buscando en la tabla de áreas bajo la curva normal tipificada (de 0 a z) obtenemos que para z = 1.18 la probabilidad es de 0.3810 = 38.10%. Esta es P(0 < z < 1.18).

Como preguntan P(x > 150), tenemos que

P(x > 150) = 0.5 - P(0 < z < 1.18) = 0.5 - 0.3810 = 0.1190

La probabilidad de que un cliente compre más de $150 es del 11.90%.

=======

c)

x = 120

Normalizamos:

.......... _

z = (x - x) / σ

z = (120 - 128.45) / 18.26

z = -8.45 / 18.26

z ≅ -0.46

Buscando en la tabla de áreas bajo la curva normal tipificada (de 0 a z) obtenemos que para z = 0.46 la probabilidad es de 0.1772 = 17.72%. Esta es P(-0.46 < z < 0) = P(0 < z < 0.46).

Como preguntan P(x < 120), tenemos que

P(x < 120) = 0.5 - P(-0.46 < z < 0) = 0.5 - 0.1772 = 0.3228

<em><u>RESPUESTA</u></em><em><u>:</u></em><em><u> </u></em>

La probabilidad de que un cliente compre menos de $120 es del 32.28%

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