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CALCULO INVERSO
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TRABAJO N° 2
DOCENTE TITULAR: BARRETO RODRIGUEZ CARMEN ROSA
DOCENTE: QUISPE TICONA PEDRO LEONARDO
CURSO: ESTADISTICA APLICADA
NOMBRE DEL ALUMNO: SALMISTA DAVID PANCA HUMPIRI
CARRERA PROFESIONAL: CONTABILIDAD
CICLO: III
CALCULO INVERSO EN LA DISTRIBUCIÓN NORMAL ESTÁNDAR.
ESTANDARIZACIÓN.
1.- Si Ζ n (0,1), hallar los
valores –Ζο y Ζο en las siguientes áreas
a) P(Ζ ≤ Ζο) = 0.96 ⟹ Ζο = 1.75
b) P(Ζ ≤ −Ζο) = 0.0013 ⟹ −Ζο = −3.0
c) P(Ζ ≤ −Ζο) = 0.002 ⟹ −Ζο = −2.88
d) P(Ζ ≤ −Ζο) = 0.998 ⟹ −Ζο = −2.88
e) P(−Ζο ≤ Ζ ≤ Ζο) = 0.997 ⟹ −Ζο = −2.97 y 𝚭𝛐 = 2.97
f) P(−Ζο ≤ Ζ ≤ Ζο) = 0.966 ⟹ −Ζο = −2.12 y 𝚭𝛐 = 2.12
2.- Si X n (1000, 3600), Hallar:
a) P(X ≥ 1200) = 1 –P(X ≤ 1200)
= 1−P(Ζ ≤
)
= 1−P(Ζ ≤ 3.33) = 1 −0.9996 = 0.0004
b) P(X ≥ 850) =P(Ζ ≤
)
= P(Ζ ≤ − 2.5) = 0.0062
c) P(X ≥ 900) = 1 –P(X ≤ 900)
= 1−P(Ζ ≤
)
= 1−P(Ζ ≤ −1.67) = 1 −0.0475 = 0.9525
d) P(850 ≤ X ≤ 940)=P
≤ Ζ ≤
= P( −2.5 ≤ Ζ ≤ −1) = P(−1) –P(−2.5)
= 0.1587 – 0.0062 = 0.1525
e) P(X ≥ 1150) = 1 –P(X ≤ 1150)
= 1−P(Ζ ≤
)
= 1−P(Ζ ≤ 2.5) = 1 −0.9938 = 0.0062
f) P(820 ≤ X ≤ 880) = P
≤ Ζ ≤
= P( −3 ≤ Ζ ≤ −2) = P(−2) –P(−3)
= 0.0228 – 0.0013 = 0.0215
g) P(800 ≤ X ≤ 1200) = P
≤ Ζ ≤
= P( −3.33 ≤ Ζ ≤ 3.33) = P(3.33) –P(−3.33)
= 0.9996 – 0.0004 = 0.9992
h) P(1120 ≤ X ≤ 1180)= P
≤ Ζ ≤
= P( 2 ≤ Ζ ≤ 3) = P(3) –P(2)
= 0.9987 – 0.9772 = 0.0215
i) P(1000 ≤ X ≤ 1250)= P
≤ Ζ ≤
= P( 0 ≤ Ζ ≤ 4.17) = P(4.17) – P(0)
= 1 – 0.5 = 0.5
j) P(950 ≤ X ≤ 1200)= P
≤ Ζ ≤
= P( −0.83 ≤ Ζ ≤ 3.33) = P(3.33) –P(−0.83)
= 0.9996 – 0.2033 = 0.7963