Question

Blood type AB is the rarest blood type, occurring in only 4% of the population in the United States. In Australia, only 1.5% of the population has blood type AB. Suppose a random sample of 50 U.S. residents and 40 Australians is obtained. Consider the random variables described below:

X: the number of US residents (out of 50) with blood type AB.
Y: the number of Australians (out of 40) with blood type AB.
Z: the total number of individuals (out of 90) with blood type AB.

Which of the following is true about the random variables X, Y, and Z?

a. X is a binomial random variable with n = 50 and p = 0.04
b. Y is a binomial random variable with n = 40 and p = 0.015
c. Z is a binomial random variable with n = 90 and p = 0.055

Answer 1

Answer:

a. X is a binomial random variable with n = 50 and p = 0.04

b. Y is a binomial random variable with n = 40 and p = 0.015

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

X: the number of US residents (out of 50) with blood type AB.

Blood type AB is the rarest blood type, occurring in only 4% of the population in the United States

This means that $p = 0.04, n = 50$

Y: the number of Australians (out of 40) with blood type AB.

In Australia, only 1.5% of the population has blood type AB.

This means that $p = 0.015, n = 40$

Z: the total number of individuals (out of 90) with blood type AB.

Here

$n = 90, p = 0.04*\frac{50}{90} + 0.015*\frac{40}{90} = 0.0289$

Which of the following is true about the random variables X, Y, and Z?

Options a and b are true, while c is false.