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##### MATH302 -Midterm Exam

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Question;Question 1 of 25The mode is best described as theA.same as the averageB.middle observationC.most frequently occurring valueD.50th percentilePart 2 of 9 -Question 2 of 25A jar contains four white marbles, five red marbles, and six black marbles. If a marble is selected at random, find theprobability that it is white or black.A.2/15B.2/3C.1/5D.2/5Question 3 of 25The formal way to revise probabilities based on new information is to use:A.common sense probabilitiesB.unilateral probabilitiesC.complementary probabilitiesD.conditional probabilitiesPart 3 of 9 -Question 4 of 25If a gambler rolls two dice and gets a sum of 10, he wins \$10, and if he gets a sum ofthree, he wins \$20. The cost to play the game is \$5. What is the expectation of thisgame?A.\$3.06B.-\$2.78C.-\$3.06D.\$2.78Question 5 of 25Suppose that 50 identical batteries are being tested. After 8 hours of continuous use, assume that a given battery is still operating with a probability of 0.70 and has failed with a probability of 0.30.What is the probability that between 25 and 30 batteries (inclusive) will last at least 8 hours?A.0.7792B.0.1186C.0.0839D.0.9169Question 6 of 25A drug is reported to benefit 40% of the patients who take it. If 6 patients take the drug, what is the probability that 4 or more patients will benefit?A.0.667B.0.179C.0.138D.0.862Part 4 of 9 -Question 7 of 25Find the probability, P(Z < 0.17), using the standard normal distribution.A..8300B..4325C..5675D..0675Question 8 of 25Given that Z is a standard normal variable, the value z for which P(Z < z) = 0.2580 isA.-0.65B.-0.70C.0.242D.0.758Question 9 of 25The average height of flowering cherry trees in a nursery is 11 feet. If the heights are normally distributed with a standard deviation of 1.6, find the probability that a randomly selected cherry tree in this nursery is less than 13 feet tall.A. 0.67B.0.89C.0.95D.0.11Question 10 of 25If Z is a standard normal random variable, the area between z = 0.0 and z =1.30 is 0.4032, while the area between z = 0.0 and z = 1.50 is 0.4332. What is the area between z = -1.30 and z = 1.50?A.0.8364B.0.0668C.0.0968D.0.0300Question 11 of 25Given that the random variable X is normally distributed with a mean of 80 and a standard deviation of 10, P(85 < X 1.50) = 0.9332TrueFalse

Paper#61096 | Written in 18-Jul-2015

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