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Midterm 2A1 SP10 Solutions

# An advertising executive is studying the television

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Problems 31-35. An advertising executive is studying the television (TV) viewing habits of conventional married couples. On the basis of past records, she determined that husbands watch TV 60% of the time. She also determined that when the husband is watching TV, the wife is also watching 40% of the time, and when the husband is not watching TV, the wife is watching TV 30% of the time. H – Husband Watching TV W – Wife Watching TV P(H)=0.6 P(W|H)=0.4 P(W|H C )=0.3 31. Make a two-way table of this situation and fill in all its probabilities, including those in the margins. P(W and H) = P(W|H)P(H) = (0.4)(0.6) = 0.24 P(W and H C ) = P(W|H C )P(H C ) = (0.3)(1-0.6) = 0.12 P(W C and H) = P(H) – P(W and H) = 0.6 – 0.24 = 0.36 P(W C and H C ) = P(H C ) – P(W and H C ) = 0.4 – 0.12 = 0.28 W W C H 0.24 0.3 6 0. 6 H C 0.12 0.28 0. 4 0.3 6 0.6 4 1 32. Find the overall percentage of time the wives watches TV. P(W) = P(W and H) + P(W and H C ) = P(W|H)P(H) + P(W|H C )P(H C ) = (0.4)(0.6) + (0.3)(1-0.6) = 0.24 + 0.12 = 0.36 (Could also read off of the table) Ans__ 0.36 ___ 33. What total percentage of the time do the wives and husbands watch TV together? P(W and H) = P(W|H)P(H) = (0.4)(0.6) = 0.24 (Could also read off of the table) 7

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Ans__ 0.24 ____ 8
34. Suppose the wife is watching TV. What’s the chance she is watching by herself? ) ( ) ( ) | ( W P H and W P W H P C C = ) ( ) ( ) | ( W P H P H W P C C = 36 . 0 ) 6 . 0 1 )( 3 . 0 ( - = 36 . 0 12 . 0 = (Could read these off of the table) = 0.333 Ans__ 0.333 __ 35. Suppose you walk into the room where the TV is located. Which of the four possible situations is most likely to be going on? BRIEFLY justify using probabilities. Wife and husband both watching ___________ Neither is watching ____________ Wife is watching but husband is not ____________ Husband is watching but wife is not ____ X _______ Wife and husband both watching: P(W and H) = 0.24 Neither is watching: P(W C and H C ) = 0.28 Wife is watching but husband is not: P(W and H C ) = 0.12 Husband is watching but wife is not: P(W C and H) = 0.36 Husband is watching but wife is not is the most likely situation. It has the highest probability. Problems 36-39. Suppose 70% of teenagers send text messages on their phone, 80% of them have accounts on Facebook, and 60% of teenagers have both. T – Sends Text Messages F – Has a Facebook Account P(T) = 0.7 P(F) = 0.8 P(T and F) = 0.6 36. What’s the probability that a teenager sends text messages or has an account on Facebook, but NOT both? P(T or F but not both) = P(T) + P(F) – 2P(T and F) = 0.7 + 0.8 – 2(0.6) = 0.3 ANS__ 0.3 ________ 9

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37. What’s the probability that a teenager sends text messages, has an account on Facebook, or both? P(T or F) = P(T) + P(F) – P(T and F)
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