Question

A marketing firm would like to test-market the name of a new energy drink targeted at 18- to 29-year-olds via social media. A study by the Pew Research Center found that 35% of U.S. adults (18 and older) do not use social media (Pew Research Center website, October 2015). The percentage of U.S. young adults age 30 and older is 78%. Suppose that the percentage of the U.S. adult population that is either age 18–29 or uses social media is 67.2%. a. What is the probability that a randomly selected U.S. adult uses social media? b. What is the probability that a randomly selected U.S. adult is aged 18–29? c. What is the probability that a randomly selected U.S. adult is 18–29 and a user of social media? Anderson, David R.. Essentials of Statistics for Business and Economics (p. 198). Cengage Learning. Kindle Edition.

Answer 1

Answer:

(a) The probability that a randomly selected U.S. adult uses social media is 0.35.

(b) The probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c) The probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.

Step-by-step explanation:

Denote the events as follows:

X = an US adult who does not uses social media.

Y = an US adult between the ages 18 and 29.

Z = an US adult between the ages 30 and above.

The information provided is:

P (X) = 0.35

P (Z) = 0.78

P (Y ∪ X') = 0.672

(a)

Compute the probability that a randomly selected U.S. adult uses social media as follows:

P (US adult uses social media (X')) = 1 - P (US adult so not use social media)

                                                   $=1-P(X)\\=1-0.35\\=0.65$

Thus, the probability that a randomly selected U.S. adult uses social media is 0.35.

(b)

Compute the probability that a randomly selected U.S. adult is aged 18–29 as follows:

P (Adults between 18 - 29 (Y)) = 1 - P (Adults 30 or above)

                                            $=1-P(Z)\\=1-0.78\\=0.22$

Thus, the probability that a randomly selected U.S. adult is aged 18–29 is 0.22.

(c)

Compute the probability that a randomly selected U.S. adult is 18–29 and a user of social media as follows:

P (Y ∩ X') = P (Y) + P (X') - P (Y ∪ X')

                $=0.22+0.65-0.672\\=0.198$

Thus, the probability that a randomly selected U.S. adult is 18–29 and a user of social media is 0.198.