Ex6soln - STAT 1306 INTRODUCTORY STATISTICS ANSWERS 2...

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STAT 13 EXAMPLE CLA 06 INTRODUCTORY STATISTICS SS 6 ANSWERS 1. If Z ~ N ( μ = 0 , 2 σ = 1 ) is the standard normal random variable and X ~ N ( μ = 80 , 2 = 2 10 ) then: a) P(Z<-1.37) = 0.5 - 0.4147 = 0.0853 < | | > Z -1.37 0 TABLES: Look up 1.37 in table A gives probability=0.4147 b) P(Z>-1.75) = 0.5 + 0.4599 = 0.9599 < | | > Z -1.75 0 TABLES: Look up 1.75 in table A gives probability=0.4599 1
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STANDARDIZED SCORES OR Z-SCORES σ μ = X z c) P(X>90) = ] 10 80 90 [ > Z P = P[ Z > 1] = 0.5 - 0.3413 = 0.1587 < | | > X 80 90 < | | > Z 0 1.0 TABLES: Look up 1.0 in table A gives z=0.3413 2
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2. Let X = the scores for the reading comprehension test X ~ N( μ = 80 , = 16) 2 σ a) P[ 78 < X < 85] = ] 4 80 85 4 80 78 [ < < Z P = P[-0.5 < Z < 1.25] = 0.1915 + 0.3944 = 0.5859 < | | | > X 78 80 85 < | | | > Z -0.5 0 1.25 Look up 0.5 in table A, gives prob=0.1915 Look up 1.25 in table A, gives prob=0.3944 3
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3. If Z ~ N ( μ = 0 , 2 σ = 1 ) is the standard normal random variable and X ~ N ( μ = 80 , 2 10 = ) then: 2 a) P(0 < Z < z) = 0.379 therefore z = 1.17 from table B < | | > Z 0 1.17 b) P( |Z| < z) = 0.95 P(-z < Z < z) = 0.95 therefore z = 1.96 Look up 0.475 in table B < | | | > Z -1.96 0 1.96 4
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P(Z >z)=0.9 therefore z = -1.282 Look up 0.40 in table B < | | > Z -1.282 0 d) P(X<x) = 0.95 95 . 0 ] 10 80 10 80 [ = < x X P P[ Z< 10 80 x ] = 0.95 Now for P[Z<z] = 0.95 table B give z = 1.645 (Look up 0.45) Thus, 10 80 x = 1.645 Therefore, x = 1.645 x 10 + 80 = 96.45 < | | > X 80 x=? < | | > Z
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Ex6soln - STAT 1306 INTRODUCTORY STATISTICS ANSWERS 2...

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