Distribucion normal y binomial ejercicios resueltos

Solución de la guía de distribución binomial y normal

1) n= 5
p=0.50

a) p(x=2) = p(x≤2)-p(x≤1)

=0.5000-0.1875

=0.3125

b) p(x≥1) = 1-p(x≤0)

=1-0.0312

=0.9688

c) p(x>3)= 1-p(x≤3)

=1-0.8125

=0.1875

d) p(x=0)= (5C0)(0.50)⁰(0.50)⁵=0.0312

2) n=10
p=1/6

a) p(x=0)=(10C0)(1/6)⁰ (5/6)¹⁰

=0.1615

b) p(x>8)=p(x≥9)

=0

c) p(x≥3)=1-p(x≤2)
=1-(p(x=0)+p(x=1)+p(x=2)
=1-((10C0)(1/6)⁰ (5/6)¹⁰+(10C1)(1/6)¹ (5/6)⁹+(10C2)(1/6)²(5/6)⁸
=1-(0.1615+0.3230+0.2907)
=1-0.7752=0.2248

d) p(x=10)= 0

3) n=3
p=0.50

p(x=3)= (3C3) (0.50)³(0.50)⁰

=0.1250

P(x=2)= (3C2) (0.50)²(0.50)¹

=0.3750

P(x=1)= (3C1) (0.50)¹(0.50)²

=0.3750

P(x=0)= (3C0) (0.50)⁰(0.50)³

=0.1250

Comprobación= 0.1250+0.3750+0.3750+0.1250= 1

μ= (3) (0.50)= 1.5

σ²= (3) (0.50)(0.50)= 0.754) n=4
p=0.543

a) p(x=2)= (4C2) (0.543)²(0.457)²

=0.3694

b) p(x=4)= (4C4) (0.543)⁴(0.457)⁰
=0.0869

c) p(x≥1)= 1- p(x≤0)

=1- (4C0) (0.543)⁰ (0.457)⁴

=1-0.0436

=0.9564

d) p(x=0)= 0.0436

5) n=7
p=0.15

a) p(x=7)= p(x≤7)-p(x≤6)

=0

b) p(x=2)= p(≤2)-p(x≤1)= 0.9262-0.7166

=0.2096

c) p(x=0)= 0.3206

d) p(x=0)= 0.3206

6) n=25
p=0.25

a) p(x=25)= (25C25) (0.25)²⁵(0.75)⁰

= 0

b) p(x≥22)= 0

c) p(x=23)= 0

d) p(x=0)= 0.008

7) n=5
p=0.98

a) p(x=5)= (5C5) (0.98)⁵(0.02)⁰

=0.9039

b) p(x=0)= 0

c) p(x≥2)= p(x=2)+p(x=3)+p(x=4)+p(x=5)= (5C2) (0.98)²(0.02)³+ (5C3)(0.98)³(0.02)²+(5C4)(0.98)⁴(0.02)¹+0.9039

= 0.000077+0.003764+0.0922+0.9039

=0.9998

d) p(x>2)= p(x=3)+p(x=4)+p(x=5)

= 0.0037+0.0922+0.9039

= 0.9993

8) n=15
p=1/6

a) p(x=10)= (15C10) (1/6)¹⁰(5/6)⁵

=0.00002

b) p(x≥2)= 1-p(x≤1)

=1-{p(x=0)+p(x=1)}

= 1-{(15C0)(1/6)⁰(5/6)¹⁵+ (15C1)(1/6)¹(5/6)¹⁴}

= 1-(0.0649+0.1947)

= 1-0.2596

= 0.7404

c) p(x=0)= (15C0)(1/6)⁰(5/6)¹⁵
= 0.0649

d) p(x>10)= p(x=11)+p(x=12)+p(x=13)+p(x=14)+p(x=15)

= (15C11) (1/6)¹¹(5/6)⁴+ (15C12)(1/6)¹²(5/6)³+(15C13)(1/6)¹³(5/6)²

+(15C14)(1/6)¹⁴(5/6)¹+ (15C15) (1/6)¹⁵(5/6)⁰

= 0

9) n=5
p=0.39

a) p(x=2)= p(x≤2)-p(x≤1)

= 0.6997-0.3545

= 0.3452

b) p(x>3)= 1- p(x≤3)

= 1- 0.9204

= 0.0796

c) p(x=0)= 0.0845

d) p(x=3)= p(x≤3)- p(x≤2)

= 0.9204- 0.6997

= 0.2207

10) Encontrar la media y la varianza delos ejercicios anteriores:

10.1) μ= (5) (0.50)= 2.5

σ²= (5) (0.50) (0.50)= 1.25

10.2) μ= (10) (1/6)= 1.67

σ²= (10) (1/6) (5/6)= 1.39

10.3) μ= (3) (0.50)= 1.5

σ²= (3) (0.50) (0.50)= 0.75

10.4) μ= (4) (0.543)= 2.172

σ²= (4) (0.543) (0.457)= 0.992

10.5) μ= (7) (0.15)= 1.05

σ²= (7) (0.15) (0.85)= 0.892510.6) μ= (25) (0.25)= 6.25

σ²= (25) (0.25) (0.75)= 4.6875

10.7) μ= (5) (0.98)= 4.9

σ²= (5) (0.98) (0.02)= 0.098

10.8) μ= (15) (1/6)= 2.5

σ²= (15) (1/6) (5/6)= 2.08

10.9) μ= (5) (0.39)= 1.95

σ²= (5) (0.39) (0.61)= 1.1895

11) Con la ayuda de la tabla de distribución N(0.1), halla:

a) P(z≤1)= 0.8413

b) P(z≥0.56)=...