est ad is tic as
TRANSCRIPT
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EJEMPLO DE LA DISTRIBUCIÓN DE LA PROBABILIDAD BINOMIAL
Una fábrica de medicamentos realiza pruebas clínicas con 100 nuevos fármacos potenciales. Cerca del 20% de las sustancias que alcanzan esta etapa reciben finalmente la aprobación para su venta ¿Cuál es la probabilidad de que se aprueben al menos 15 de los 100 medicamentos? suponga que se satisfacen las hipótesis de la distribución binomial, y utilice una aproximación normal con corrección por continuidad.
Solución formula
p ( x=k ) [ nk ] pk .qn−k
La media (valor esperado) de y es μ=100 (0.2 )=20 ; la desviación estándar es
σ √100 (0.2 ) (0.8 )=4.0¿
¿la probabilidad buscada es 15 o mas medicamentos se aprueben. Como
y= está incluido, la corrección por continuidad consiste en tomar el evento como y ≥14.5
p¿
Un examen consta de 10 preguntas al as que hay que contestar si o no suponiendo que a las personas que se le aplica no saben contestar a ninguna de las preguntas y,en consecuencia, contestan al azar, hallar :
a) probabilidad de obtener 5 aciertos
b) probabilidad de obtener algún acierto
c) probabilidad de obtener al menos 5 aciertos
Es una distribución binomial, la persona solo puede acertar o fallar la pregunta
Suceso A= (éxito)=acertar la pregunta→ p=p ( A )0.5
SucesoA=noacertar la pregunta→q=p ( A )=0.5
Distribución binomial de parámetros n=10 , p=0.5→B (10 ;0.5 )
A) Probabilidad de obtener 5 aciertos
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Obtener exactamente 5 aciertos k=5, aplicamos la formula:
P ( X=K )=[ nk ]Pk .qnk →k=5n=10 p=0.5q=0.5→P ( x=5 )=[ 105 ] . (0.5 )5 . (0.5 )10−5
[ nk ]= n
k ! (n−k )!numeros combinatorios→ [ 105 ]=¿
P(x=5)=¿
b) Probabilidad de obtener algún acierto
p ( x≥1 )=p ( x=1 )+ p ( x=2 )+ p ( x=3 )+ p ( x=4 )+p (x=5 )+p (x=6 )+ p ( x=7 )+ p ( x=8 )+ p ( x=9 )+ p (x=10)
El suceso “obtener algún acierto “es el suceso contrario a “no obtener ningún acierto “
P ( X=0 )=[ 100 ] . (0.5 ) . (0.5 )10=0.00100
px (≥1 )=1−p ( x=0 )→ p ( x≥1 )=−0.00100=0.999
c) probabilidad de obtener al menos 5 aciertos acertar 5 o mas
p ( x≥5 )=p ( x=5 )+ p ( x=6 )+ p ( x=7 )+ p ( X=8 )+ p ( x=9 )+ p (x=10 )
p ( x≥5 )=0.2461+0.2051+0.1172+0.0439+0.0098+0.0010=06231
2 la probabilidad de que un estudiante obtenga el título de licenciado en farmacia es 0.3
Hallar la probabilidad de que un grupo de siete estudiantes matriculados en primer curso finalice la carrera
a) ninguno de los siete finalice la carrerab) b)finalicen todosc) al menos dos acaben la carrera d) hallar la media y la desviación típica del número de alumnos que acabaran la carrera
A=” obtener el tiulo”→ p=p ( A )=0.3
A= “ no obtener el tiulo “→q=P ( A )=1−0.3=0.7→B=7 ;0.3
A) ninguno de los siete finalice la carrera x=0
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P ( x=k )=[ nk ] pk .qn−k →k=0n=7 p=0.3q=0.7→ p ( x=0 )=[ 70 ] . (0.3 ) . (0.7 )7−0=0.0824
b) finalicen todos x=7
p(x=k)=[ nk ] pk .qn−k →k=7n=3 p=0.3q=0.7→ p=( x=7 )=[ 77 ] . (0.3 )7. (0.7 )0=0.0002
c) al menos dos terminan la carrerax≥2
calculamos la probabilidad del suceso contrario , probabilidad que no termine ninguno mas la probabilidad de que termine uno
P ( x≥2 )=1−¿
La probabilidad de que termine ninguno :P ( X=1 )=[ 71 ] . (0.3 )1. (0.7 )6=0.2471
P ( X ≥2 )=1−[ P ( X=0 )+P ( X=1 )→P ( X ≥2 )=1−[0.0824+0.2471 ] ]=06705
D) Media y desviación típica
Media μ=n . p=7.0,3=2.1
Desviación típica σ √n . p .q √7.0,3 .0,7=1.2124
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DISTRIBUCIÓN DE PROBABILIDADES HIPERGEOMETRICA
Consideramos tomar una muestra de 10 de las 87 cuentas de una compañía de las 87, 13 tenían errores. Encuentre p(2 cuentas incorrectas en la muestra.
Solución
Tenemos N=87 , n=10 , Nϵ=13 y , por lo tanto ,Nf =74 ;queremos p ( y=2 ) .
pγ (2 )(132 )
7410
−2
❑
❑❑ ( 8710 )
=1,175,600,000,0004000,800,000,000
=.294
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