Statistics And Probability

Question

63% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 32 owned dogs are randomly selected, find the probability that

a. Exactly 21 of them are spayed or neutered.

b. At most 19 of them are spayed or neutered.

c. At least 20 of them are spayed or neutered.

d. Between 15 and 20 (including 15 and 20) of them are spayed or neutered.

Answer 1

X is a binomial random variable. These questions can be calculated using a binomial function. I'll explain using the TI-84 function

Density function

P(X=12) = binompdf(32, 0.63, 12) = 0.0020

Cumulative Distribution Function

P(X<=19) = binomcdf(32, 0.63, 19) = 0.3989

P(X>=20) = 1 - P(X<=19) = 1- 0.3989 = 0.6011

P(15 <= X <= 20) = P(X <= 20) - P(X <= 15)

= binomcdf(32, 0.63, 20) - binomcdf(32, 0.63, 15) = 0.5431 - 0.0459 = 0.4972