Question

Left-handedness occurs in about 12% of all Americans. Males are slightly more likely than females to be left-handed, with 13% of males and 11% of females being left-handed. Suppose a random sample of 80 females and 100 males is chosen.
-Let X be the number of males (out of the 100) who are left-handed.

-Let Y be the number of females (out of the 80) who are left-handed.

-Let Z be the total number of left-handed individuals in the sample (males and females together).

Which of the following is true regarding the random variables X and Y?

a)Both X and Y can be well approximated by normal random variables.
b)Only X can be well approximated by a normal random variable.
c)Only Y can be well approximated by a normal random variable.
d)Neither X nor Y can be well approximated by a normal random variable.

Answer 1

Answer:

a)Both X and Y can be well approximated by normal random variables.

Step-by-step explanation:

For each individual, there are only two possible outcomes. Either they are right-handed, or they are left-handed. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability:

Probability of exactly x sucesses on n repeated trials, with probability p.

The binomial probability can be well approximated by normal random variables, using the expected value $E(X) = np$ and the standard deviation $\sigma = \sqrt{np(1-p)}$

Let X be the number of males (out of the 100) who are left-handed.

$n = 100$ and $p = 0.13$. Can be well approximated.

Let Y be the number of females (out of the 80) who are left-handed.

$n = 80$ and $p = 0.11$. Can be well approximated.

The correct answer is

a)Both X and Y can be well approximated by normal random variables.