2 2 2 2 1 σ σ π σ x x x e g 5 5 2 1 x x 2 2 1 σ

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2 2 2 ) ( , 2 1 σ σ π σ X x X e G = " 5 ' 5 2 1 = = x X " 2 2 1 = σ (a) higher than 5’10” means Δ x = 4 ½ ” above the mean height, 8 . 1 " 2 / " 4 / 2 1 2 1 = = Δ = σ x t
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t = 1.8. The value found in the table is 92.81. What percentage of the women are taller than 5’10” ? 92.81 3.6 The answer is (100 – 92.81)/2 3.6. 3.6% out of 2000 amounts to 72
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The cutoff at 5’10” gives a t-value of 1.8 and 3.6% of women eligible to join. Now we want to have 7.2% of women being eligible. What do we do? 072 . 0 2 1 2 2 2 ) ( , = = h X x h X e G σ σ π σ We want to have 856 . 0 ) 072 . 0 2 100 ( , = = + σ σ σ t X t X X G
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All the club has to do is lowering the bar by 1”, from 5’10” to 5’9”! 85.6 7.2 t -? 856 . 0 ) 072 . 0 2 100 ( ) ( , = = + x G t X t X X σ σ σ " 9 ' 5 " 2 46 . 1 " 5 ' 5 46 . 1 2 1 2 1 + = + = = σ t x h t
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Typical situation - a sample (set) of N numeric values of the same parameter What are the things we usually want to know? Statistical analysis of data N x N x x x x i N = + + + = ... 2 1 = = N i i x x x N 1 2 2 ) ( 1 σ N x x x x x x ,..., , , , 4 3 2 1 = 1) The mean value: 2) The variance and standard deviation = = N i i x x x N 1 2 ) ( 1 σ
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