Applied Mathematics">
Taller N5
Taller N5
Taller N5
Solución:
X = número de horas
µ = 5040 hrs
σ = 720 hrs
P ( x< a )=0.90
x−5040
(
P Z<
720 )
=0.90
( x−5040
720 )
=1.282
X =5963.04
x−80
(
P Z<
12 )
=0.25
( x−80
12 )
=−0.67
X =71. 96
Respuesta: el 25% de los alumnos con menor fluidez verbal los separa un
71,96 de puntuación.
b) P ( x ≥ a )=0.25
1−P ( Z ≤ a )=0.25
0.75=P ( Z ≤ a )
0.67= ( x−80
12 )
X =88. 04
10−µ
=−1.036
σ
10+1.036 σ =µ … .(1)
P ( x ≥ 13 )=0.08
13−µ
1−P(Z ≤ )=0.08
σ
13−µ
=−1.405
σ
13+1.405 σ =µ … .(1)
σ = -8.13
µ = 1.58
P ( x< 4 )=0.02
4−µ
P( Z< )=0.02
0.04
4−µ
=−0.082
0.04
µ=4.082
Solución:
µ=4
σ=2
P( 4−a ≤ x ≤ 4+ a)=0.5934
P ( x ≤ 4+ a )−P ( x ≤ 4−a)=0.5934
4+(a−4 ) 4−(a−4)
(
P Z≤
2 )
−P(Z ≤
2
)=0.5934
( a2 )−P (Z ≥ a2 )=0.5934
P Z≤
( a2 )−[1−P ( Z ≤ a2 )]=0.5934
P Z≤
( a2 )−[1−P ( Z ≤ a2 )]=0.5934
P Z≤
2P Z≤( a2 )−1=0.5934
2P Z≤( a2 )=1 .5934
( a2 )=0.7967
P Z≤
a
=0.83
2
a=1. 66