Practice Midterm 1 Solutions-1

Practice Midterm 1 Solutions-1 - Practice Midterm I...

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Practice Midterm I Solutions 1. (1) D (2) A (3) B (4) C (5) B 2. a) mean = (15+17+99+11+15+16)/6 = 28.83 median: first arrange numbers from least to greatest: 11 15 15 16 17 99 Now, take middle two numbers, average them. So, median = (15+16)/2 = 15.5 mode: 15 , since it is the number occurring the most frequently b) variance = [(15-28.83)^2+(17-28.83)^2+(99-28.83)^2+(11-28.83)^2+(15- 28.83)^2+(16-28.83)^2]/(6-1) = 5,928.83/5 = 1,185.77 c) Looking at ordered data in increasing order: 11 15 15 16 17 99, divide data into two sets: 11 15 15, and 16 17 99 for data to the left and to the right of the median. The median of the first set, 15 , is the First Quartile, and the median of the second set, 17 , is the Third Quartile. The Interquartile Range, IQR, is equal to 17-15 = 2 . 3. Quarts SD Frequency Relative Frequency Cumulative Frequency 0 --3 4 0.20 0.20 4 --7 5 5 0.25 0.45 8 --11 6 6 0.30 0.75 12 -- 15 3 2 0.10 0.85 16 --19 2 3 0.15 1.00 ----------- ----------------- 20 1.00 4. a) Let event T = event that Tom passes the test Let event D = event that Dick passes the test Since T and D are independent events, then P(T and D) = P(T)*P(D). P(T) = 0.7, P(D) = 0.8, P(not T) = 1-0.7 = 0.3, P(not D) = 1-0.8 = 0.2
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Practice Midterm 1 Solutions-1 - Practice Midterm I...

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