Stat HW2

# Stat HW2 - Statistics 285 Solutions to HW#2 Problems 3.2 a...

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Statistics 285 Solutions to HW #2 Problems 3.2 a. This is a Venn Diagram. b. If the sample points are equally likely, then P (1) = P (2) = P (3) = ⋅ ⋅ ⋅ = P (10) = 1 10 Therefore, P ( A ) = P (4) + P (5) + P (6) = 1 1 1 3 10 10 10 10 + + = = .3 P ( B ) = P (6) + P (7) = 1 1 2 10 10 10 + = = .2 c. P ( A ) = P (4) + P (5) + P (6) = 1 1 3 5 20 20 20 20 + + = = .25 P ( B ) = P (6) + P (7) = 3 3 6 20 20 20 + = = .3 3.9 a. The sample points of this experiment correspond to each of the 6 possible colors of the M&M’s. Let Br =brown, Y =yellow, R =red, Bl =blue, O =orange, G =green. The six sample points are:{ Br , Y , R , Bl , O , and G} b. From the problem, the probabilities of selecting each color are: P ( Br ) = 0.13, P ( Y ) = 0.14, P ( R ) = 0.13, P ( Bl ) = 0.24, P (O) = 0.2, P ( G ) = 0.16 c. The probability that the selected M&M is brown is P(Br) = 0.13 d. The probability that the selected M&M is red, green or yellow is: P ( R or G or Y ) = P ( R ) + P ( G ) + P ( Y ) = 0.13 + 0.16 + 0.14 = 0.43 e. P (not Bl ) = P ( R ) + P ( G ) + P ( Y ) + P ( Br ) + P ( O ) = 0.13 + 0.16 + 0.14 + 0.13 + 0.20 = 0.76 3.12 Define the following event: B : {Postal worker was assaulted on the job in the past year}

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P ( B ) = 600 .05 12,000 = 3.19 Since one would be selecting 3 stocks from 15 without replacement, the total number of ways to select the 3 stocks would be a combination of 15 things taken 3 at a time. The number of ways would be 455 )! 3 15 ( ! 3 ! 15 3 15 = - = 3.25 a. The number of ways the 5 commissioners can vote is 2(2)(2)(2)(2) = 2 5 = 32 (Each of the 5 commissioners has 2 choices for his/her vote – For or Against.) b. Let F denote a vote ‘For’ and A denote a vote ‘Against’. The 32 sample points would be: FFFFF FFFFA FFFAF FFAFF FAFFF AFFFF FFFAA FFAFA FAFFA AFFFA FFAAF FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA FAAAF AFFAA AFAFA AFAAF AAFFA AAFAF AAAFF FAAAA AFAAA AAFAA AAAFA AAAAF AAAAA Each of the sample points should be equally likely. Thus, each would have a probability of 1/32. c. The sample points that result in a 2-2 split for the other 4 commissioners are: FFAAF FAFAF AFFAF FAAFF AFAFF AAFFF FFAAA FAFAA FAAFA AFFAA AFAFA AAFFA There are 12 sample points.
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