Distribución de La Probabilidad Binomial... EJERCICIOS SOLUCIONADOS

ESTADÍSTICA GENERAL-I DISTRIBUCIÓN DE LA PROBABILIDAD BINOMIAL 1. Para una distribui!n bin"#ia$ "n n % 1& ' (% ).*+, nuntr a/

P ( x = 8 )

b/

P ( x > 4 )

/

P ( x ≤ 10 )

Solución :

()

 x n− x  P ( x  x )= n  p q  x

 p=0.45 q =1− 0.45 ⟹ q =0.55

a/

(

P  x = 8

)=

( ) 12 8

8

8

( 0.45 ) ( 0.55 )

 P ( x  x = 8 )=( 495 ) ( 0.45 ) ( 0.55 )

b/

4

↪

4

 P ( x = 8 )=0.076

P ( x  x > 4 ) = [1 − P ( x  x ≤ 4 ) ]

{

 P ( x  x > 4 ) = 1− [ P  P ( x =0 ) + P ( x =1 ) + P ( x =2 )+ P ( x =3 )+ P ( x = 4 ) ]

{ [((

 P ( x  x > 4 ) = 1 −

/

)

0 12 12 ( 0.45 ) ( 0.55 ) 0

) (( ) +

}

) (( )

1 11 12 ( 0.45 ) ( 0.55 ) 1

+

2 10 12 ( 0.45 ) ( 0.55 ) 2

) (( ) +

3 9 12 ( 0.45 ) ( 0.55 ) 3

) (( ) +

4 12 ( 0.45 ) ( 4

1− ¿

 P ( x  x ≤ 10 ) =¿

 P ( x  x ≤ 10 )={1 −[ P  P ( x =11 )+ P ( x =12 ) ] }

{ [((

 P ( x  x ≤ 10 )= 1−

)

11 1 12 ( 0.45 ) ( 0.55 ) 11

) (( ) +

12 0 12 ( 0.45 ) ( 0.55 ) 12

)]}=

0.999

MAYKOL JHOEL

SOLUCIONES

ESTADÍSTICA GENERAL-I

&. Enuntra $a #dia, 0ariana ' $a ds0iai!n st2ndar d $as si3uints distribui"ns bin"#ia$s a/

n =

14, p % ).*).

b/

n

% 1),

 p

% ).5+.

/

n

% &&,

 p

% ).1+.

d/

n

% 6+),  p %

).7). /

n

% 58,

 p

%

).)+. S"$ui!n Sa  μ= media 2

σ  = varianza 2

σ  = desviación est á ndar

a/

n =16 , p =0.40 , q =1− p= 0.60

{

μ =np ⟹ μ=( 16 ) ( 0.40 )=6.40 σ  =npq ⟹ σ  = (16 ) ( 0.40 ) ( 0.60 )=3.84 2

2

σ = √ npq ⟹ σ =√ ( 16 ) ( 0.40 ) ( 0.60 )=1.96

b/

n =10 , p =0.75 , q=1 − p=0.25

{

μ =np ⟹ μ= (10 ) ( 0.75 ) = 7.50 σ  =npq ⟹ σ  =( 10 ) ( 0.75 ) ( 0.25 )= 1.88 2

2

σ = √ npq ⟹ σ =√ ( 10 ) ( 0.75 ) ( 0.25 ) = 1.37

/

n =22 , p = 0.15 ,q =1 − p =0.85

{

μ =np ⟹ μ=( 22 ) ( 0.15 )= 3.30 σ  =npq ⟹ σ  =( 22 ) ( 0.15 ) ( 0.85 )= 2.81 2

2

σ = √ npq ⟹ σ =√ ( 22 ) ( 0.15 ) ( 0.85 ) = 1.67

d/

n =350 , p =0.90 , q=1 − p=0.10

{

μ =np ⟹ μ=( 350 ) ( 0.90 ) = 315.00 σ  =npq ⟹ σ  = ( 350 ) ( 0.90 ) ( 0.10 )=31.50 2

2

σ = √ npq ⟹ σ =√ ( 350 ) ( 0.90 ) ( 0.10 )=5.61

/

n =78 , p =0.05 , q=1 − p=0.95

MAYKOL JHOEL

SOLUCIONES

ESTADÍSTICA GENERAL-I

{

μ =np ⟹ μ =( 78 ) ( 0.05 )= 3.90 2 2 σ  =npq ⟹ σ  =( 78 ) ( 0.05 ) ( 0.95 )= 3.71 σ = √ npq ⟹ σ =√ ( 78 ) ( 0.05 ) ( 0.95 ) = 1.92

6. E$ 9$ti#" s"nd" ("$:ti" nai"na$ india ;u $a (r"babi$idad d ;u iudadan"s $3id"s a$ aar san "nsr0ad"rs s d ).++< d ;u san $ibra$s s d ).6), ' d ;u st=n ntr una ' "tra "rintai!n s ).1+. Su("n3a ;u stas (r"babi$idads s"n >atas ' rs("nda a $as si3uints (r3untas ?2$u$"s #anua$s/ r@ridas a un 3ru(" d 1) iudadan"s s$i"nad"s d #anra a$at"ria a/ Cu2$ s $a (r"babi$idad d ;u uatr" san $ibra$s b/ Cu2$ s $a (r"babi$idad d ;u nin3un" sa "nsr0ad"r / Cu2$ s $a (r"babi$idad d ;u d"s st=n ntr una ' "tra "rintai!n d/ Cu2$ s $a (r"babi$idad d ;u a$ #n"s "" san $ibra$s S"$ui!n

a/

n =10, p= 0.30, q =1 − p =0.70 ⟹ P

b/

4

6

4

n =10, p= 0.55, q =1 − p =0.45 ⟹ P

/

( )

( x = 4 ) = 10 ( 0.30 ) ( 0.70 ) = 0.2001

( )

( x =0 )=

10 0

0

10

( 0.55 ) ( 0.45 ) =0.0003

n = 10, p= 0.15, q =1− p =0.85

( )

2 8  P ( x =2 ) = 10 ( 0.15 ) ( 0.85 ) = 0.2759

⟹

d/

2

n = 10, p= 0.30, q =1 − p =0.70  P ( x ≥ 8 )= P ( x =8 ) + P ( x =9 ) + P ( x =10 )

⟹

⟹

(( )

10 ( 0.30 )8 ( 0.70 )2 8

) (( ) +

10 ( 0.30 )9 ( 0.70 )1 9

) (( ) +

10 ( 0.30 )10 ( 0.70 )0 10

)=

0.0016

*. ar$' Da0ids"n, dirt"r d "ntr"$ d a$idad d $a "#(a:a d aut"#!0i$s F'"t" M"t"r, s nuntra ra$iand" su r0isi!n #nsua$ d trans#isi"ns aut"#2tias. En $ (r"di#int", s rtiran 1) trans#isi"ns d $a (i$a d "#("nnts ' s $s r0isa n busa d d@t"s d @abriai!n. A $" $ar3" d$ ti#(", s!$" $ & d $as trans#isi"ns tinn d@t"s ?su("n3a ;u $"s d@t"s s (rsntan d #anra ind(ndint n di@rnts trans#isi"ns/. MAYKOL JHOEL

SOLUCIONES

ESTADÍSTICA GENERAL-I

a/ Cu2$ s $a (r"babi$idad d ;u $a #ustra d ar$' "ntn3a #2s d d"s

trans#isi"ns "n d@t"s d @2bria b/ Cu2$ s $a (r"babi$idad d ;u nin3una d $as trans#isi"ns $3idas tn3a d@t"s d @2bria S"$ui!n Sa  x = transmisiones defectuosas n =10  p=0.02 q =1− p =0.98

(

P  x > 2

a/

)= 1− P ( x ≤ 2 )=[ 1−( P ( x =0 )+ P ( x =1 ) + P ( x =2 ) ) ]

 P ( x > 2 )= 1−

{[( )

10 ( 0.02 )0 ( 0.98 )10 0

] [( ) +

10 ( 0.02 )1 ( 0.98 )9 1

] [( ) +

10 ( 0.02 )2 ( 0.98 )8 2

]}

 P ( x > 2)= 0.00086

( )

0 10 P ( x = 0 )= 10 ( 0.02 ) ( 0.98 ) =0.81707

b/

0

+. Diana Bru" s $a a$a$dsa d una iudad 3rand. H$ti#a#nt, s a stad" (r"u(and" ara d $a

("sibi$idad d ;u 3rands antidads d

(rs"nas ;u "bran $ s3ur" d ds#($" n ra$idad tn3an un

traba" n

srt". Sus asistnts sti#an ;u *) d $"s bnJiari"s d$ s3ur" d ds#($" ntra n sta at3"r:a, (r" $a s"ra Bru" n" st2 "n0nida. L (id a un" d sus a'udants ;u a3a una in0sti3ai!n d &) bnJiari"s d$ s3ur" t"#ad"s a$ aar. a/ Si $"s asistnts d $a a$a$dsa tinn ra!n, u2$ s $a (r"babi$idad d ;u $"s indi0idu"s in0sti3ad"s tn3an un #($" b/ Si $"s asistnts d $a a$a$dsa st2n n $" "rrt", u2$ s $a (r"babi$idad d ;u s!$" in" d $"s indi0idu"s in0sti3ad"s tn3an traba" / Si $"s asistnts d $a a$a$dsa st2n n $" "rrt", u2$ s $a (r"babi$idad d ;u ;uin " #2s d $"s S"$ui!n

indi0idu"s in0sti3ad"s tn3an traba"

n =20,  p = 0.4, q =0.6

a/ b/

( ) ( = )=( )(

20 0 P ( x = 20 )= 2 0 ( 0.4 ) ( 0.6 ) =¿  ).)))))))11 20

P  x

5

20 5

0.4

5

15

) ( 0.6 ) =0.074647020

MAYKOL JHOEL

SOLUCIONES

ESTADÍSTICA GENERAL-I

/

(( )

) (( )

) (( )

) (( )

) (( )

) ((

15 5 16 4 17 3 18 2 19 1  P ( x ≥ 15 )= 20 ( 0.4 ) ( 0.6 ) + 20 ( 0.4 ) ( 0.6 ) + 20 ( 0.4 ) ( 0.6 ) + 20 ( 0.4 ) ( 0.6 ) + 20 ( 0.4 ) ( 0.6 ) +

15

16

17

18

19

 P ( x ≥ 15 )=0.001294494 + 0.000269686 + 0.000042304 + 0.000004700 + 0.000000330+ 0.000000011

 P ( x ≥ 15 ) =0.001611525

MAYKOL JHOEL

SOLUCIONES