Question

27. 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 per centage 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?

Answer 1

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Let event A adult uses social media @ =P(A) = l-PCA) - ".P(A) = = - 0.35 = 0.65 65 % (Aus). 6 let event Be adult is aged 18-29 event c = adult is aged 30 f older Then from given information we have P (c) = 78.1. ☆ PC BUA) = 57.24. – ☺ also, PC (BU BC) 2A) = 35% o P(BrAC) + P( CnA = 354 -③ { B & C arez from 0. disjoint = P(B) + P(A) - P(ABB) = 67-2 y. =) P (B)o-P(ARB) = 67.24. – 65%. = 2.24. = P(BAAC) = 2.2%. B & C are

using this in ® we get, PCCNA") = 35 t. - 2.2t. P(COAC) = 32.84. on P (CNA) = P(c) - P(COAC) = 784.- 32-8 y. op CCNA) - 45.2.1. ---☺ Now, from 35 y. ( 18 f older) don't of us adults use social media 3) 657. use it. a PC BUC)NA) = 654 o PC And ) +P(ANC) -65%. P(ARB) =(65-45.2)). 19.8t (from ③].

P(B) = 2-2 y.t 19.8y. P(B) = 224. (Ans) ©. - P(BOA) = 19.84: ' (us) [from