Question

85% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 38 owned dogs are randomly selected, find the probability that a. Exactly 31 of them are spayed or neutered. b. At most 34 of them are spayed or neutered. c. At least 30 of them are spayed or neutered. d. Between 28 and 36 (including 28 and 36) of them are spayed or neutered.

Answer 1

n=	38	p=	0.8500
here mean of distribution=μ=np=		32.3
and standard deviation σ=sqrt(np(1-p))=		2.2011
for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

a)

Exactly 31 of them are spayed or neutered:

probability =

P(30.5<X<31.5)

=

P(-0.82<Z<-0.36)=

0.3594-0.2061=

0.1533

( please try 0.1399 if this comes wrong)

b)

P At most 34 of them are spayed or neutered )

probability =

P(X<34.5)

=

P(Z<1)=

0.8413

(try 0.8421 if this comes wrong)

c)

At least 30 of them are spayed or neutered :

probability =

P(X>29.5)

=

P(Z>-1.27)=

1-P(Z<-1.27)=

1-0.1020=

0.8980

(try 0.8943 if this comes wrong)

d)

P(28 <=X<=36)=0.9573

(try 0.9633 if this comes wrong)