ESTADÍSTICA GENERAL-I DISTRIBUCIÓN DE LA PROBABILIDAD BINOMIAL 1. Para una distribui!n bin"#ia$ "n n % 1& ' (% ).*+, nuntr 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
&. Enuntra $a #dia, 0ariana ' $a ds0iai!n st2ndar d $as si3uints distribui"ns bin"#ia$s a/
n =
14, p % ).*).
b/
n
% 1),
p
% ).5+.
/
n
% &&,
p
% ).1+.
d/
n
% 6+), p %
).7). /
n
% 58,
p
%
).)+. S"$ui!n Sa μ= media 2
σ = varianza 2
σ = desviación est á ndar
a/
n =16 , p =0.40 , q =1− p= 0.60
{
μ =np ⟹ μ=( 16 ) ( 0.40 )=6.40 σ =npq ⟹ σ = (16 ) ( 0.40 ) ( 0.60 )=3.84 2
2
σ = √ npq ⟹ σ =√ ( 16 ) ( 0.40 ) ( 0.60 )=1.96
b/
n =10 , p =0.75 , q=1 − p=0.25
{
μ =np ⟹ μ= (10 ) ( 0.75 ) = 7.50 σ =npq ⟹ σ =( 10 ) ( 0.75 ) ( 0.25 )= 1.88 2
2
σ = √ npq ⟹ σ =√ ( 10 ) ( 0.75 ) ( 0.25 ) = 1.37
/
n =22 , p = 0.15 ,q =1 − p =0.85
{
μ =np ⟹ μ=( 22 ) ( 0.15 )= 3.30 σ =npq ⟹ σ =( 22 ) ( 0.15 ) ( 0.85 )= 2.81 2
2
σ = √ npq ⟹ σ =√ ( 22 ) ( 0.15 ) ( 0.85 ) = 1.67
d/
n =350 , p =0.90 , q=1 − p=0.10
{
μ =np ⟹ μ=( 350 ) ( 0.90 ) = 315.00 σ =npq ⟹ σ = ( 350 ) ( 0.90 ) ( 0.10 )=31.50 2
2
σ = √ npq ⟹ σ =√ ( 350 ) ( 0.90 ) ( 0.10 )=5.61
/
n =78 , p =0.05 , q=1 − p=0.95
MAYKOL JHOEL
SOLUCIONES
ESTADÍSTICA GENERAL-I
{
μ =np ⟹ μ =( 78 ) ( 0.05 )= 3.90 2 2 σ =npq ⟹ σ =( 78 ) ( 0.05 ) ( 0.95 )= 3.71 σ = √ npq ⟹ σ =√ ( 78 ) ( 0.05 ) ( 0.95 ) = 1.92
6. E$ 9$ti#" s"nd" ("$:ti" nai"na$ india ;u $a (r"babi$idad d ;u iudadan"s $3id"s a$ aar san "nsr0ad"rs s d ).++< d ;u san $ibra$s s d ).6), ' d ;u st=n ntr una ' "tra "rintai!n s ).1+. Su("n3a ;u stas (r"babi$idads s"n >atas ' rs("nda a $as si3uints (r3untas ?2$u$"s #anua$s/ r@ridas a un 3ru(" d 1) iudadan"s s$i"nad"s d #anra a$at"ria a/ Cu2$ s $a (r"babi$idad d ;u uatr" san $ibra$s b/ Cu2$ s $a (r"babi$idad d ;u nin3un" sa "nsr0ad"r / Cu2$ s $a (r"babi$idad d ;u d"s st=n ntr una ' "tra "rintai!n d/ Cu2$ s $a (r"babi$idad d ;u a$ #n"s "" san $ibra$s S"$ui!n
a/
n =10, p= 0.30, q =1 − p =0.70 ⟹ P
b/
4
6
4
n =10, p= 0.55, q =1 − p =0.45 ⟹ P
/
( )
( x = 4 ) = 10 ( 0.30 ) ( 0.70 ) = 0.2001
( )
( x =0 )=
10 0
0
10
( 0.55 ) ( 0.45 ) =0.0003
n = 10, p= 0.15, q =1− p =0.85
( )
2 8 P ( x =2 ) = 10 ( 0.15 ) ( 0.85 ) = 0.2759
⟹
d/
2
n = 10, p= 0.30, q =1 − p =0.70 P ( x ≥ 8 )= P ( x =8 ) + P ( x =9 ) + P ( x =10 )
⟹
⟹
(( )
10 ( 0.30 )8 ( 0.70 )2 8
) (( ) +
10 ( 0.30 )9 ( 0.70 )1 9
) (( ) +
10 ( 0.30 )10 ( 0.70 )0 10
)=
0.0016
*. ar$' Da0ids"n, dirt"r d "ntr"$ d a$idad d $a "#(a:a d aut"#!0i$s F'"t" M"t"r, s nuntra ra$iand" su r0isi!n #nsua$ d trans#isi"ns aut"#2tias. En $ (r"di#int", s rtiran 1) trans#isi"ns d $a (i$a d "#("nnts ' s $s r0isa n busa d d@t"s d @abriai!n. A $" $ar3" d$ ti#(", s!$" $ & d $as trans#isi"ns tinn d@t"s ?su("n3a ;u $"s d@t"s s (rsntan d #anra ind(ndint n di@rnts trans#isi"ns/. MAYKOL JHOEL
SOLUCIONES
ESTADÍSTICA GENERAL-I
a/ Cu2$ s $a (r"babi$idad d ;u $a #ustra d ar$' "ntn3a #2s d d"s
trans#isi"ns "n d@t"s d @2bria b/ Cu2$ s $a (r"babi$idad d ;u nin3una d $as trans#isi"ns $3idas tn3a d@t"s d @2bria S"$ui!n Sa 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$dsa d una iudad 3rand. H$ti#a#nt, s a stad" (r"u(and" ara d $a
("sibi$idad d ;u 3rands antidads d
(rs"nas ;u "bran $ s3ur" d ds#($" n ra$idad tn3an un
traba" n
srt". Sus asistnts sti#an ;u *) d $"s bnJiari"s d$ s3ur" d ds#($" ntra n sta at3"r:a, (r" $a s"ra Bru" n" st2 "n0nida. L (id a un" d sus a'udants ;u a3a una in0sti3ai!n d &) bnJiari"s d$ s3ur" t"#ad"s a$ aar. a/ Si $"s asistnts d $a a$a$dsa tinn ra!n, u2$ s $a (r"babi$idad d ;u $"s indi0idu"s in0sti3ad"s tn3an un #($" b/ Si $"s asistnts d $a a$a$dsa st2n n $" "rrt", u2$ s $a (r"babi$idad d ;u s!$" in" d $"s indi0idu"s in0sti3ad"s tn3an traba" / Si $"s asistnts d $a a$a$dsa st2n n $" "rrt", u2$ s $a (r"babi$idad d ;u ;uin " #2s d $"s S"$ui!n
indi0idu"s in0sti3ad"s tn3an 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