quiz2_solutions - 55 = 1 2(27 37 1 2(41 77 = 34 57(c 45...

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Name: Math 1011– Intro to Statistics Spring 2011 QUIZ 2 — Normal Distribution 1/31/11 1. Suppose a set of test scores are well approximated by a normal distribution with mean μ = 54 and standard deviation σ = 11. (a) What percentage of scores were above 70? (b) What percentage of scores were between 50 and 60? (c) What percentage of scores were between 45 and 50? (d) What was the 90th percentile, i.e. the score at which 90% scored that score or less? Solution: (a) 70 converts to 1.45 in standard units. Area(1 . 45 , ) = 1 2 (Area( -∞ , ) - Area( - 1 . 45 , 1 . 45)) = 1 2 (100% - 85 . 29%) = 7 . 36% (b) 50 converts to - 0 . 36 and 60 converts to 0.55 in stnadard units Area( - 0 . 36 , 0 . 55) = Area( - 0 . 36 , 0) + Area(0 , 0 . 55) = 1 2 Area( - 0 .
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Unformatted text preview: , . 55) = 1 2 (27 . 37%) + 1 2 (41 . 77%) = 34 . 57% (c) 45 converts to-. 82 and 50 converts to-. 36 in standard units. Area(-. 82 ,-. 36) = Area(-. 82 , 0)-Area(-. 36 , 0) = 1 2 Area(-. 82 , . 82)-1 2 Area(-. 36 , . 36) = 1 2 (57 . 63%)-1 2 (27 . 37%) = 15 . 13% (d) We want to Fnd x such that 90% = Area(-∞ , x ). 90% = Area(-∞ , x ) = Area(-∞ , 0) + Area(0 , x ) = Area(-∞ , 0) + 1 2 Area(-x, x ) = 50% + 1 2 Area(-x, x ) 80% = Area(-x, x ) Now consulting our table, we see x ≈ 1 . 30, and then converting this from standard units back to our test score gives 68.3....
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